calculus - How to show that $lim_{nto infty } left(frac{(1 + frac{1}{n^2})^{n^2}}{e}right)^n = 1$?

I need to find the limit:
$$\lim_{n\to \infty } \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n$$
So I know that the limit is $1$.
Using Squeeze theorem
$$? \leq \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n \leq \left(\frac{e}{e}\right)^n \rightarrow\ 1 $$
What should be instead $?$ ? Is it possible to solve in another way?
Unfortunately, I can't use L'Hôpital Rule or Series Expansion in this task.

Answer

Note that:

$$\left(\frac{(1 + \frac{1}{n^2})^{n^2}}{(1 + \frac{1}{n^2})^{n^2+1}}\right)^n=\left(1 + \frac{1}{n^2}\right)^{-n}\leq \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n \leq \left(\frac{e}{e}\right)^n=1$$

Since:

$$\left(1 + \frac{1}{n^2}\right)^{-n}\to 1$$

By Squeeze Theorem:

$$\lim_{n\to \infty } \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n=1$$