sequences and series - Convergence of $sum_{ kinmathbb{N} } left( frac{lambda^k}{k!} right)^n$

We know that $\sum_{ k\in\mathbb{N} } \frac{\lambda^k}{k!} = e^\lambda$. I'm interested in the convergence of

$$S^{(n)}=\sum_{ k\in\mathbb{N} } \left( \frac{\lambda^k}{k!} \right)^n$$
for some value of $n$.
The series general term $a_k= \left( \frac{\lambda^k}{k!} \right)^n$ converges for $k\geq 1$. Applying the comparison test of this series with exponential series
$$\lim_{k\rightarrow+\infty} \frac{\left( \frac{\lambda^k}{k!} \right)^n}{\frac{\lambda^k}{k!}} = 0$$
, we may conclude that the sequence converges; i.e. the series converges.
Numerically i tested the behavior for $n\rightarrow+\infty$ of $S^{(n)}$ and the series converges to $2$.


Now, if there is no mistake in what I wrote before, I wan't to know if there is any way to calculate the value of convergence $S$? Is there any relation between $S$ and $e$ or $e^\lambda$?

Answer

Let

$$S^{(n)}(\lambda)=\sum_{k=0}^\infty\frac{\lambda^{nk}}{(k!)^n}.$$
Then
$$\lim_{n\to\infty}S^{(n)}(\lambda)=\begin{cases} 0 & \text{if }0\le\lambda<1,\\ 2 & \text{if }0\le\lambda=1,\\ \infty &\text{if }\lambda>1. \end{cases}$$
Let's prove it. If $0\le\lambda<1$ then
$$1\le S^{(n)}(\lambda)=1+\sum_{k=1}^\infty\frac{\lambda^{nk}}{(k!)^n}\le1+\sum_{k=1}^\infty\lambda^{nk}=1+\frac{\lambda^n}{1-\lambda^n}.$$
If $\lambda>1$ then
$$\le S^{(n)}(\lambda)\ge 1+\lambda^n.$$
Finally, since $k!\le2^k$ if $k\ge4$, we have
$$2\le S^{(n)}(1)\le2+\frac{1}{2^n}+\frac{1}{6^n}+\sum_{k=4}^\infty\frac{1}{2^{nk}}=2+\frac{1}{2^n}+\frac{1}{6^n}+\frac{2^{-4n}}{1-2^{-n}}.$$