A trigonometric integral (guessed from a combinatorics formula) In class, I defined the binomial coefficient using an integral:

$$\binom{n}{k} = \displaystyle \int_0^{2\pi}\dfrac{dt}{2\pi} e^{-ikt}(1+e^{it})^n.$$

I succeeded in demonstrating many standard properties of the binomial coefficient directly using integration: Pascal's identity, Vandermonde identity, Hockey stick. But I could not show that $$\sum_{k=0}^n \binom{n}{k}=2^n.$$
It turns out I have to show the following:
$$\int_{0}^{2\pi}\dfrac{dt}{2\pi}\dfrac{\sin\left(\dfrac{(n+1)t}{2}\right)}{\sin\dfrac{t}2}\cos^n\dfrac{t}{2}=1$$ 
I do not know how to perform this integration! I need help. It is better if the solution did not involve contour integration.
 A: Hint 
Notice that inside the integral we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}$$ which is equal to $D_{\frac n2}$ where $D_n$ denotes the Dirichlet kernel. 
Hence using it's properties we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}=D_{\frac n2}=1+2\sum_{r=1}^{\frac n2} \cos (rt) $$
Hope you can continue further 
A: Using mathematical induction. Base case $n=0$, ok!
Assume that is true for an arbitrary natural number $n$. Since 
$$ \sum_{k=0}^{n+1} \binom{n+1}{k} = \sum_{k=0}^n \binom{n+1}{k} + \binom{n+1}{n+1} $$
Using your definition, $\binom{n+1}{n+1}$ equals to $1$, then using Pascal's identity
$$ \binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1} \text{ for all } 1\leqslant k\leqslant n $$
You get the following :
$$ \sum_{k=0}^{n+1} \binom{n+1}{k} = 2^n + 2^n - 1 +1 = 2^{n+1}$$
A: There is a mistake in my presentation that eludes me, but the essence is the following.
Using 
$$\sum_{k=1}^{\infty} p^{k} \, \sin(k \, x) = \frac{p \, \sin(x)}{1 - 2 \, p \, \cos(x) + p^{2}}$$
then let $S_{n}$ be the desired summation to be evaluated to obtain:
\begin{align}
S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \\
&= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin\left(\frac{t}{2}\right)} \, dt \\
&= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin(t)} \, dt 
\end{align}
Now,
\begin{align}
\sum_{n=0}^{\infty} S_{n} &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=0}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \sin\left(\frac{(n+1) \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\
&= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \sin\left(\frac{n \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\
&= \frac{1}{2 \pi} \, \int_{0}^{2 \pi} \frac{2 \cos\left(\frac{t}{2}\right) \sin\left(\frac{t}{2}\right)}{1 - 4 \cos^{2}\left(\frac{t}{2}\right) + 4 \cos^{2}\left(\frac{t}{2}\right)} \, \frac{dt}{\sin(t)} \\
&= \frac{1}{2\pi} \, \int_{0}^{2 \pi} dt = 1 \\
&= - (-1) = - \frac{1}{1 - 2} = - \sum_{n=0}^{\infty} 2^{n}. 
\end{align}
This yields
$$\sum_{n=0}^{\infty} \binom{n}{k} = - 2^{n}.$$
As stated, the mistake that eludes me is the sign error.
