$\lim_{n\rightarrow\infty} n\left|e^\frac{i2\pi}{n}-1\right|$ $$\lim_{n\to\infty}n\left|e^\frac{i2\pi}n-1\right|$$
I know that it must be $2\pi$ since it's just approximating the perimeter of the circle, but I can't do the limit algebraically.
I tried using the De Moivre formula: $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, didn't work.
Any tips?
Edit: solved using @xpaul advice since it was the thing I was doing but I missed a 2 in the expansion of $(1-\cos(\frac{\pi}{n}))^2$.
Also solved it using Robert Israel hint because it was very clever. 
 A: Noting
\begin{eqnarray*} 
1-e^{\frac{2\pi i}{n}}|&=&\bigg|1-\cos(\frac{2\pi}{n})-i\sin(\frac{2\pi}{n})\bigg|\\
&=&\sqrt{\bigg[1-\cos(\frac{2\pi}{n})\bigg]^2+\sin^2(\frac{2\pi}{n})}\\
&=&\sqrt{2-2\cos(\frac{2\pi}{n})}\\
&=&2\sin(\frac{\pi}{n})
\end{eqnarray*}
one has
$$\lim_{n\rightarrow\infty} n\left|e^\frac{i2\pi}{n}-1\right|=\lim_{n\to\infty}2n\sin(\frac{\pi}{n})=2\pi. $$
Here
$$ \lim_{x\to0}\frac{\sin x}{x}=1. $$
A: Since $\exp\frac{2\pi i}{n}-1=2i\sin\frac{\pi}{n}\exp\frac{\pi i}{n}$, your limit is $\lim_{n\to\infty}2n\sin\frac{\pi}{n}=2\pi$. Of course, you could also just use the approximation $\exp z-1\in z+O(z^2)$.
A: Define $$f(x) := e^{i2\pi x}.$$ Then we have 
$$\lim_{n\to\infty}n\left|e^\frac{i2\pi}n-1\right| = \lim_{n\to\infty} \left|\frac{f(1/n) - f(0)}{1/n - 0}\right| = |f'(0)| = 2\pi.$$
A: Alternatively:
$$\lim_{n\to\infty}n\left|e^\frac{i2\pi}n-1\right|=\left|\lim_{n\to\infty}\frac{e^\frac{i2\pi}n-1}{\frac1n}\right|=|2\pi i|=\sqrt{(2\pi)^2}=2\pi.$$
Note:
$$\lim_{x\to 0} \frac{e^{ax}-1}{x}=a.$$
