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. 

Solution
Applying $\textbf{L'Hospital's Rule}$, we have $$\begin{align*}\lim_{x \to +\infty} \left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]&=\lim_{x \to +\infty}\dfrac{\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\&=\lim_{x \to +\infty}\dfrac{1}{\sin\left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\cos \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \sin\left(\dfrac{\pi}{2}+\dfrac{2}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \cos\left(\dfrac{2}{x}\right)}\\&=2. \end{align*}$$ Hence, $$\begin{align*}\lim\limits_{n\to \infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right)&=\lim\limits_{x\to +\infty}\tan^x\left(\frac{\pi}{4}+\frac{1}{x}\right)\\&=\lim\limits_{x\to +\infty} \exp\left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\exp\left[\lim_{x \to \infty}x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\mathbb{e}^2.\end{align*}$$
