Q:

Need help showing $\left(\frac{n^2 - 2n + 1}{n^2-4n+2}\right)^n \to e^2$

I'm trying to work out this limit:
$$
\lim_{n \to \infty} \left(\frac{n^2 - 2n + 1}{n^2-4n+2}\right)^n = e^2
$$
 I have managed to show that it is equal to: $$\lim_{n \to \infty}\left (1+\frac{2n-1}{n^2-4n+2}\right)^n$$ but I have no idea how to continue from here, help would be appreciated. 

A:

HINT
We have that
$$\left(\frac{n^2 - 2n + 1}{n^2-4n+2}\right)^n =\left(\frac{n^2 - 4n + 2+2n-1}{n^2-4n+2}\right)^n =\left(1+\frac{2n-1}{n^2-4n+2}\right)^n = \left[\left(1+\frac{2n-1}{n^2-4n+2}\right)^{\frac{n^2-4n+2}{2n-1}}\right]^\frac{n(2n-1)}{n^2-4n+2}$$

