Consider the following, where $n\geq 7$ is a natural number:
$$\frac{1}{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}.$$
I am convinced that there should be an upper bound of the form
$$\frac{1}{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}\leq e^{-c(n)},$$
for $c(n)$ some positive function of $n$ but I am failing dismally to derive one.
Any help would be gratefully appreciated.
Edit: I think $c(n)\approx 4.689$ works and suffices for my needs. This is independent of $n$.
Note that we can write
$$\begin{align} \frac1{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}&=e^{-\pi^2}\left(1+\frac{e^{\pi^2/n^2}-1}{2}\right)^{n^2}\\\\ &\le e^{-c(n)} \end{align}$$
where $c(n)$ is given by
$$c(n)\le\pi^2-n^2\log\left(1+\frac{e^{\pi^2/n^2}-1}{2}\right)$$
The asymptotic expansion of $c(n)$ is given by
$$\begin{align} c(n)&=\pi^2-n^2 \log\left(1+\frac{(\pi^2/n^2+\frac12(\pi^2/n^2)^2+\frac16(\pi^2/n^2)^3+O\left(1/n^8\right)}{2}\right)\\\\ &=\pi^2-n^2\left(\frac{\pi^2}{n^2}+\frac{\pi^4}{8n^4}+O(1/n^8)\right)\\\\ &=\frac{\pi^2}{2}-\frac{\pi^4}{8n^2}+O\left(\frac1{n^6}\right)\\\\ &=c_{\text{approx}}+O\left(\frac1{n^6}\right) \end{align}$$
where $c_{\text{approx}}=\frac{\pi^2}{2}-\frac{\pi^4}{8n^2}$.
For $n=7$, we have
$$\begin{align}c(7) &\approx 4.68672854698357\\ c_{\text{approx}}(7) &=4.68630962137631 \end{align}$$