Good upper bound on $f(n)$ This is an elementary question. I'm trying to understand the function (a discrete),$$f(n) = \left(\displaystyle \dfrac{\sum_{k=0}^n \frac{a^k}{k!}}{\sum_{k=0}^{n-1}\frac{a^k}{k!}}\right)^n\,$$ where $a>0$ is a constant.
I used WolframAlpha to check the behavior of a graph of a function similar to this and it turns out it's a bell curve (like the Binomial Distribution), which means that this's neither monotonically increasing nor decreasing.So, there must be a local maximum but I don't think there's a proper way to figure out the exact maximum value. Anyways, I'm interested in a good upper bound on $f(n)$.
Also, for a given $a>0$, I noticed that $f(n)\to 1$ as $n\to \infty.$ It can be proved that $f(n)^{1/n} \to 1$ as $n\to \infty$ but not sure how to show that for $f(n)$.
Any help would be apreciated.

Let $g(n):=\sum_{k=0}^n\frac{a^k}{k!}$ so that $g(n)=g(n-1)+\frac{a^n}{n!}$ and $\lim_{n\to\infty}g(n)=e^a$. Then
$$f(n)=\left(\frac{g(n)}{g(n-1)}\right)^n
=\left(\frac{g(n-1)+\tfrac{a^n}{n!}}{g(n-1)}\right)^n
=\left(1+\frac{a^n}{n!g(n-1)}\right)^n.$$
This shows that for  $h(n)=\frac{a^n}{(n-1)!g(n-1)}$ we have
$$\lim_{n\to\infty}f(n)
=\lim_{n\to\infty}\left(1+\frac{h(n)}{n}\right)^n
=\lim_{n\to\infty}e^{h(n)}=1.$$
because $\lim_{n\to\infty}h(n)=0$.
