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}}.
$$