Computing limit of $(1+1/n)^{n^2}$

By Bernoulli's Inequality, $$\left(1+\frac{1}{n}\right)^{n^2}\geq 1+n$$ so the result follows trivially $\square$

There is a slight problem when you say "the whole things is $e^n$": what do you mean? Surely you know you cannot chose when to let the $n$s go to infinity, you have to make them large "at the same pace".

Take the logarithm of that, and see what happens. That is, you have $$x_n=\left(1+\frac 1 n\right)^{n^2}$$ Then $$\log x_n=n^2\log\left(1+\frac 1 n\right)$$

and $$\log\left(1+\frac 1n\right)=\frac 1n -\frac{1}{2n^2}+O\left(\frac 1{n^3}\right)$$

Alternatively, since $$\left(1+\frac 1 n\right)^n\to e $$

there exists $N$ such that whenever $n\geq N$, we have $$\left(1+\frac 1 n\right)^n\geq \frac{e}2\text{ (Why?) }$$

We have $\dfrac e2>1$ since $e>2$ and then $$x_n\geq r^n$$ whenever $n>N$ with $r>1$.