How to show that $\lim_{n\to \infty } \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n = 1$? 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. 
 A: If $$A=\lim_{n\to \infty } \left(\dfrac{\left(1 + \dfrac1{n^2}\right)^{n^2}}e\right)^n$$
$$\ln A=\lim_{n\to \infty }n^3\ln\left(1+\dfrac1{n^2}\right)-n$$
Set $1/n=h$ and 
use Are all limits solvable without L'Hôpital Rule or Series Expansion
or L'Hôpital Rule 
or Series Expansion
A: Note that for $x>0$ the logarithm function satisfies the fundamental inequality (prove it yourself, it is easy) $$\frac{x^{2}}{1+x^{2}}\leq \log(1+x^{2})\leq x^{2}$$ or $$\frac{-x}{1+x^2}\leq \frac{\log(1+x^2)}{x^{3}}-\frac{1}{x}\leq 0$$ Putting $x=1/n$ we get $$-\frac{n} {1+n^2}\leq n^3\log(1+n^{-2})-n\leq 0$$ And using Squeeze theorem we have $n^3\log(1+n^{-2})-n\to 0$. On exponentiating we see that the limit in question is $1$.
A: 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:
$$\lim_{n\to \infty } \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n=1$$
