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.
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
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 :
$$ \sum_{k=0}^{n+1} \binom{n+1}{k} = 2^n + 2^n - 1 +1 = 2^{n+1}$$