Background
(You can skip this part, but maybe you find it interesting.)
Is $ \displaystyle f_1(x,v) = \sum_{n=0}^{\infty} \frac{x^n}{(n!)^v } > 0 $ for all real $x$ and $0<v<1$ ?
Lets start simple and take the case $v = \frac{1}{2}$ and notice the inverse ratio of Taylor coefficients is $(a_{n+1}/a_n)^{-1} = \sqrt n$. Probably the simplest way to prove positivity is now to write
$$ \int_{0}^{1} (1-t^n)\left(\log\frac{1}{t}\right)^{-3/2} \, \frac{dt}{t} = c\sqrt{n}$$
with some fixed positive $c$. Notice that the integral converges and the integrand is positive, and make the change of variable $t^n\to t$. We conclude,by the additive property and the identity $x^m t^m = (xt)^m$, that
$$ \int_{0}^{1} (f(x)-f(xt))\left(\log \frac{1}{t} \right)^{-3/2} \, \frac{dt}{t} = cxf(x). $$
If $x$ is the largest zero of $f$ (which must be negative), then by plugging it in, we get $0$ on the right and a negative number on the left, which is a clear contradiction. Thus, crossing the $x$-axis is not possible.
Of course, there is nothing special about $v = \frac{1}{2}$. Any power $v$ between $0$ and $1$ works just as well because the analogue integral still converges. For instance,
\begin{gather*} \int_{0}^{1}(1-t^n)\left(\log \frac{1}{t}\right)^{-5/4} \, \frac{dt}{t} = 4 \Gamma(3/4) n^{1/4} \qquad\text{and} \\ \int_{0}^{1} (f(x)-f(xt)) \left(\log \frac{1}{t}\right)^{-5/4} \, \frac{dt}{t} = cxf(x). \end{gather*}
This method cries out for a generalization and a deeper understanding.
Notice also that the limit of $f(x,v)$ as $x \to -\infty$ is zero and the function is strictly increasing. Also, partially because of that, the function is estimated (and proven) to go to zero at rate $O(n^{-v})$ (not so easy to prove?) and to $+\infty$ at rate $O(\exp(x^{1/v}))$.
So this is where the idea came from. There are other ways to prove positivity, but I really like this one. (A conjecture is that all similar questions with confirmation of positivity need to be reduced or be reducible to the above, or be a sequence of compositions that themselves are.)
So, we focus on the integral transform $M(t(n)) = m(t)$ for a given $t(n)$, such that
\begin{equation} \int_{0}^{1} (1-t^n) m(t) \, dt = t(n) \label{int_tf}\tag{*} \end{equation}
is valid for integer $n>0$ with $m(t)$ being positive for $0<t<1$.
However, it seems
$$ \lim_{x \to -\infty} \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} = -1 $$
AND even
$$ 1 + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} > 0 $$
This made me wonder about the integral transform...
EDIT 1
After some testing it seems more like
$$ \lim_{x \to -\infty} \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} = -1$$
AND
$$0 \leq C + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $$
Where $C = 1.036055393...$ is the best possible constant.
This seems to suggest that the integral transform does not exist since the function does not start from the empty product or equivalently
$$1 + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} > 0$$
Is false. But there has to be a better more direct way to show it right?
Question
Consider the integral transform $M(t(n)) = m(t)$ as in \eqref{int_tf}. I wonder if there is a solution of
$$ M((n+1)^n) = m(t),$$
and if so, what it is? So, we want to find strict positive $m(t)$ such that
$$\int_{0}^{1} (1-t^n) m(t) \, dt= \frac{(n+1)^n}{n^{n-1}}$$
valid for integer $n>0$.
It seems like a reasonable question. And it seems closely related to Laplace and similar transforms or the Bernstein theorem. Yet I'm not sure how to do that.
I think I can restate the problem with Lambert-W functions or so, but then again I doubt that would help much.
(Notice $n$-th derivatives of Lambert-W are $\frac{(-n)^n}{n!}$ and you already see the similarity.)
Any ideas?
EDIT 2 (A similar conjecture)
Not essential but beyond a comment perhaps.
How to prove that for all real $x$ we have:
$$ 0 \leq f(x) = A + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^{\frac{n}{2}}} $$
Where $A = 1$ is the best possible constant.
Somewhat related or similar: Conjecture about $ f(x,v) = 1 + \sum_{n=1}^\infty {x^n \over t(v n) } > 0 $ for all real $x$
