Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module. 
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning? 
Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$, 
it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,
and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.
Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.
Edit: Here's another attempt I've made.
After using addition formula for tangent, we get 
$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$
so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have 
$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$
I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts. 

Using the addition angle formula for the tangent function we can write
$$\tan^n\left(\frac{\pi}{4}+\frac1n\right)=\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n$$
Note that $\tan(x)=x+O(x^3)$.  Then, we have 
$$\begin{align}
\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n&=\left(\frac{1+\frac1n +O\left(\frac1{n^3}\right)}{1-\frac1n +O\left(\frac1{n^3}\right)}\right)^n\\\\
&=\left(\frac{1+\frac1n }{1-\frac1n }\right)^n\left(\frac{1+\frac{O\left(\frac1{n^3}\right)}{1+\frac1n}}{1 +\frac{O\left(\frac1{n^3}\right)}{1-\frac1n }}\right)^n\\\\
&\to \frac {e}{e^{-1}}\,\frac{1}{1}\,\,\text{as}\,\,n\to \infty\\\\
&=e^2
\end{align}$$
And we are done!
