(The presentation that follows is a very light introduction. For more austere results, consult Manuscript 1).

Let's begin with some notation:

Let n in N be a positive integer. We define as usual:

n! = 1*2*3*...*(n-1)*n.

Now we will define a new sequence of iterated factorials for n in N, as follows:

hf(n,k) = {n! iff k=1, [hf(n,k-1)]!, iff k > 1}. (1)

hf(n,2) = [n!]!,
hf(n,5) = [[[[n!]!]!]!]!, etc.

Where no confusion shall arise, we will use:
hf(n,k) = n!!...!, (k-"!"s).

Some examples with Maple:

3!! = 720

4!! = 620448401733239439360000

3!!! =~ .2601218944 * 10¹⁷⁴⁷

etc.

We also use the hyperpower function defined here as:
F(x,y,k) = {x^y iff k =1, x^F(x,k-1), iff k > 1}, so:

F(x,1,1) = x,
F(x,1,2) = ²x = x^x,
F(x,y,2) = x^xy
...
F(x,1,k) = ^kx = x^xx...x. (k-x's).

The comparison property of divisibility:
a | b and b <>0, => |a| <= |b|.  (1)
Proof:
In any standard Number Theory book.

Corollary:
a | b and a, b > 0,  => a<=b.

A couple of immediate facts about the sequence hf(n,k):

Assume that the lemma holds for k in N, k>1:
That is: m<=n <=> hf(m, k) | hf(n,k).

From the induction hypothesis:
m<=n, =>
hf(m,k) | hf(n,k).
From the corollary, =>
hf(m,k) <= hf(n,k).
From lemma #2, =>
[hf(m,k)]! | [hf(n,k)]!, =>
hf(m,k+1) | hf(n, k+1).

For the reverse, assume:
hf(m,k+1) | hf(n, k+1).
From the corollary, =>
hf(m,k+1) <= hf(n,k+1).
From lemma #1 (for arbitrary k), =>
m<=n,
and the lemma follows.

We will need the following two lemmas along with their corollaries:

Lemma #4:
If k >= 1 then for all n in N:
hf(n+1,k+1) - hf(n,k+1) > hf(n,k+1)*[n^{hf(n+1,k) - hf(n,k)} - 1]

Proof:
Directly or by induction. The direct proof is shorter:
By lemma #3, hf(n,k+1) | hf(n+1,k+1), =>
hf(n+1,k+1) - hf(n,k+1) =
1*2*...*hf(n+1,k) - 1*2*...*hf(n,k) =
1*2*...*hf(n,k)*[(hf(n,k)+1)*(hf(n,k)+2)*...(hf(n+1,k)) - 1] =
hf(n,k+1)*[(hf(n,k)+1)*(hf(n,k)+2)*...(hf(n+1,k)) - 1].
If k >=1, => hf(n,k) + j > n, for all j.
There are hf(n+1,k) - hf(n,k) terms in the product,
and the lemma follows.

Corollary:
If k >=1 then for all n in N:
hf(n+1,k) - hf(n,k) > ^kn = F(n,1,k)

Sketch of Proof:
Iterate the weaker inequality of lemma #4:
hf(n+1,k+1) - hf(n,k+1) > n^{hf(n+1,k) - hf(n,k)},
and use induction.

Using the corollary above, the product contains at most ^k-1n terms, and
hf(n,k-1) + j > n, for all j, =>
hf(n,k)/hf(n+1,k) < 1/n ^k-1n = 1/^kn and the lemma follows.

If k >=1, then

converges for all x.

Proof:
If k=1, we get the exp series of course.
If k>1 then:
|a(n+1)/a(n)| = |x*hf(n,k)/hf(n+1,k)|.
By lemma #5, the above tends to 0, for all x.

If k =1 then the series:

converges for |x| < 1/e. If k > 1 then it converges for all x.

Proof:
If k=1, the Ratio test gives:
lim_n->+oo|a(n+1)/a(n)| = |ex| < 1, provided |x| < 1/e.

If k > 1, the same test gives:
|a(n+1)/a(n)| = (n+1)⁽ⁿ⁺¹⁾*x/[nⁿ*(n+1)] *(n+1)*hf(n,k)/hf(n+1,k)|.
Using lemma #5:
|a(n+1)/a(n)| < |(n+1)⁽ⁿ⁺¹⁾*x/[nⁿ*(n+1)] *(n+1)/^kn|
lim_n->+oo[(n+1)⁽ⁿ⁺¹⁾*x/[nⁿ*(n+1)]] = ex, as before,
and obviously:
lim_n->+oo(n+1)/^kn = 0, for k > 1,
and the rest of the lemma follows.

Lemma #8:

If k > 2 then the series:

converges for all x.

Proof:
Using lemma #5:
|a(n+1)/a(n)| < |x^xn*[x-1]/^kn|.
For sufficiently large n, x < n, =>
|a(n+1)/a(n)| < |nⁿⁿ/^kn|=|³n/^kn|, so if k > 2, =>
lim_n->+oo|a(n+1)/a(n)| < 1,
and the lemma follows.

converges for all x.

Proof:
Similar to the proof of lemma #8.

Lemma #10:

If k >=1 then the series:

converges for x in (0,e^(1/e)].

Proof:
Using lemma #5:
|a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx * 1/^kn|.
When x is in [(1/e)^e,e^(1/e)], lim_n->+ooⁿx = e^-W(-ln(x)), by the article on Infinite Exponentials, so:
lim_n->+ooⁿ⁺¹x/ⁿx = 1, while obviously:
lim_n->+oo1/^kn = 0, for k >=1, and the lemma follows.

If on the other hand, x is in (0,(1/e)^e),
|a(n+1)/a(n)| = |ⁿ⁺¹x/ⁿx*hf(n,k)/hf(n+1,k)|, and
ⁿx is a two cycle, bounded above by 1 and below by 0.
This means that we may have problems with the odd subsequence: ^2m+1x, on the denominator.
Let's see what happens. Using lemma #5, =>
|a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx*1/^kn|, as before, and
lim_m->+oo^2mx = a < 1, with a being a solution to x^xa=a, (See Solving the Second Real Auxiliary Exponential Equation),
lim_m->+oo^2m+1x = b > 0, b < a, and with b given by x^a = b, above.
This means that:
|ⁿ⁺¹x/ⁿx|, n in N, is a two-cycle.

We check the even and odd subsequences:
lim_m->+oo|a(2m+1)/a(2m)| <= lim_m->+oo|^2m+1x/^2mx*1/^k(2m)| = |b/a*0| = 0, while:
lim_m->+oo|a(2m+2)/a(2m+1)| <= lim_m->+oo|^2m+2x/^2m+1x*1/^k(2m+1)| = |a/b*0| = 0,
and the lemma follows.

(It is non-trivial to see what happens for x in (e^(1/e), +oo). You might want to investigate yourself what the relation between k and x should be, so that lim_n->+oo|ⁿ⁺¹x/ⁿx*hf(n,k)/hf(n+1,k)| is 0, a > 0, or +oo).

Proof:
Directly or by induction. The direct proof is shorter:
By lemma #3, hf(n,k) | hf(n+1,k), and obviously:
hf(n+1,k) | hf(n+1,k+1), =>
hf(n+1,k+1) - hf(n,k+1) =
1*2*...*hf(n+1,k) - 1*2*...*hf(n,k-1) =
1*2*...*hf(n,k-1)*[(hf(n,k-1)+1)*(hf(n,k-1)+2)*...(hf(n+1,k)) - 1] =
hf(n,k)*[(hf(n,k-1)+1)*(hf(n,k-1)+2)*...(hf(n+1,k)) - 1].
If k >1, => hf(n,k-1) + j > n, for all j.
There are hf(n+1,k) - hf(n,k-1) terms in the product,
and the lemma follows.

Corollary:
If k >1 then for all n in N:
hf(n+1,k+1) - hf(n,k) > ^k+1n = F(n,1,k+1)

Proof
Iterate the weaker inequality of lemma #11:
hf(n+1,k+1) - hf(n,k) > n^{hf(n+1,k) - hf(n,k-1)},
and use induction.

Using the corollary above, the product contains at most ^kn terms, and
hf(n,k-1) + j > n, for all j, =>
hf(n,k)/hf(n+1,k)< 1/n ^kn = 1/^k+1n and the lemma follows.

Lemma #13:
The series:

converges for all x > 0.

Proof:
Using lemma #12:
|a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx*1/ⁿ⁺¹n|.
If x is in [(1/e)^e,e^(1/e)] then
lim_n->+ooⁿx = e^-W(-ln(x)), so, =>
lim_n->+oo|ⁿ⁺¹x/ⁿx| = 1, as before, and
lim_n->+oo|1/ⁿ⁺¹n| = 0,
and the result follows.
If x > e^(1/e),
for sufficiently large n, x < n, =>
ⁿ⁺¹x < ⁿ⁺¹n.
Using lemma #12, =>
|a(n+1)/a(n)| <  |1/ⁿx|,
consequently:
lim_n->+oo|a(n+1)/a(n)| = 0,
and the result follows.

If on the other hand, x is in (0,(1/e)^e),
|a(n+1)/a(n)| = |ⁿ⁺¹x/ⁿx*hf(n,n)/hf(n+1,n+1)|, and
ⁿx is again a two cycle, bounded above by 1 and below by 0.
Using lemma #12:
|a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx*1/ⁿ⁺¹n|, as before, and
lim_m->+oo^2mx = a < 1, with a being a solution to x^xa=a, (See Solving the Second Real Auxiliary Exponential Equation),
lim_m->+oo^2m+1x = b > 0, b < a, and with b given by x^a = b, above.
This means that:
|ⁿ⁺¹x/ⁿx|, n in N, is a two-cycle.

We check the even and odd subsequences:
lim_m->+oo|a(2m+1)/a(2m)| <= lim_m->+oo|^2m+1x/^2mx*1/^2m+1(2m)| = |b/a*0| = 0, while:
lim_m->+oo|a(2m+2)/a(2m+1)| <= lim_m->+oo|^2m+2x/^2m+1x*1/^2m+2(2m+1)| = |a/b*0| = 0,
and the lemma follows.

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