We have
$$\lim_{n \to \infty} \left(\tan\left(\frac{\pi^n}{2^{2n}}\right)\right)^{\frac{1}{n}}$$
It can be written as $$\lim_{n \to \infty} \left(\tan\left((\frac{\pi}{4})^n\right)\right)^{\frac{1}{n}}$$
It is in the form of $0^0$. we can write the limit as:
$$\lim_{n\to\infty}e^{\dfrac{\ln\left(\tan(\frac{\pi}4)^n\right)}n}$$
By Hopital rule, the limit of exponent is equivalent to
$$\lim_{n\to\infty}\dfrac{(\frac{\pi}4)^{n}.\ln(\frac{\pi}4).\sec^2{(\frac{\pi}4)^n}.\cot(\frac{\pi}4)^n}{1}=\lim_{n\to\infty}\dfrac{(\frac{\pi}4)^{n}.\ln(\frac{\pi}4)}{\cos{(\frac{\pi}4)^n}.\sin(\frac{\pi}4)^n}=\lim_{n\to\infty}\frac{\ln(\frac{\pi}4)}2\times\dfrac{(\frac{\pi}4)^{n}}{\sin(2(\frac{\pi}4)^n)}$$
But from here I don't know how to continue.
Best Answer
There is a simpler way to do this:
First note that $$\lim_{0<y \rightarrow 0} \frac{\tan y}{y} = 1.$$
Next notice that $\frac{\pi^n}{4^n}; n=1,2,\ldots$ is an infinite sequence of positive numbers with limit $0$. Thus, putting the above together gives
$$\lim_{n \rightarrow \infty} \left(\tan \left(\frac{\pi^n}{4^n}\right)\right)× \left( \frac{4^n}{\pi^n}\right)= 1.$$
Raising both sides to the $\frac{1}{n}$-power gives the value of $\frac{\pi}{4}$.