Limit problems and quandaries: finding $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. Find  $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. What I did is: 
$\lim_\limits{n\to \infty }{({n^2-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty } {({n^2+1-1-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty }{(1-{1+n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty}[{(1-{1+n\over n^2+1})^{ n^2+1\over 1+n}}]^{(n+10)(1+n)\over (n^2+1)}$. Denoting: ${ n^2+1\over 1+n}=t$ ($t\to \infty$ as $n\to \infty$) we get: $\lim_\limits{n\to \infty}[{(1+{(-1)\over t})^{ t}}]^{(n+10)(1+n)\over (n^2+1)}=\lim_\limits{n\to \infty}e^{-{(n+10)(1+n)\over (n^2+1)}}=e^{\lim_\limits{n\to \infty}-{(n+10)(1+n)\over (n^2+1)}}=e^{-1}$. 
My question is: how can I know it is defined and lawful? I know I can use the continuity of $e$, but I don't know if it is okay that I separated the limits that way(Because maybe the first limit is not even $e$ considering other things)
.I Would appreciate your help...

$$\frac{n^2-n}{n^2+1}=1-\frac{n+1}{n^2+1}=1-\frac1{\frac{n^2+1}{n+1}}=1-\frac1{n-\frac{n-1}{n+1}}\implies$$
$$\left(\frac{n^2-n}{n^2+1}\right)^{n+10}=\left(\frac{n^2-n}{n^2+1}\right)^{n-\frac{n-1}{n+1}}\cdot\left(\frac{n^2-n}{n^2+1}\right)^{10}\cdot\left(\frac{n^2-n}{n^2+1}\right)^{\frac{n-1}{n+1}}=$$
$$=\left(1-\frac1{n-\frac{n-1}{n+1}}\right)^{n-\frac{n-1}{n+1}}\left(\frac{n^2-n}{n^2+1}\right)^{10}\left(\frac{n^2-n}{n^2+1}\right)\left(\frac{n^2-n}{n^2+1}\right)^{-\frac2{n+1}}\xrightarrow[n\to\infty]{}e^{-1}\cdot1\cdot1\cdot1=e^{-1}$$
