How to prove that $\sum_{k=0}^{n}(-1)^k \binom {n}{k}=0 $ 
Given a positive integer $n$. How to prove that $\sum\limits_{k=0}^{n}(-1)^k \dbinom{n}{k} = 0 $ ?

I've tried using mathematical induction, then:

$$p(1)=\sum_{k=0}^{1}(-1)^k \binom {1}{k}=0 $$

And my induction hypothesis is:
$$p(n)=\sum_{k=0}^{n}(-1)^k \binom {n}{k}=0 $$
So, i need prove:
$$p(n+1)=\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}=0 $$
Let $\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}$,
$$\begin{align}
\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k} &= \sum_{k=0}^{n}(-1)^{k} \binom {n}{k}+(-1)^{k} \binom {n+1}{n+1} \\
&=0 + (-1)^{k}
\end{align}$$
What am i doing wrong?
 A: $p(n)$ is the difference between the number of subsets of $\{1,2,\ldots,n\}$ with an even number of elements and the number of subsets with an odd number of elements. Any subset of $\{1,2\ldots,n\}$ is either of the even ($E$) kind or of the odd ($O$) kind, hence in order to prove that $p(n)=0$ it is enough to show that there is a bijection between $O$ and $E$.

Given $A\subseteq\{1,2,\ldots,n\}$, let $\varphi(A)=A\cup\{1\}$ if $1\not\in A$, and let $\varphi(A)=A\setminus\{1\}$ if $1\in A$.

I leave to you to prove that $\varphi$ is a bijection between $O$ and $E$.
A: In your last line I assume you are trying to split up the sum, that is, write the last term separately.  However this does not change the terms inside the sum, so it should be
$$\sum_{k=0}^{n+1}(-1)^k\binom{n+1}k
  =\left[\sum_{k=0}^n(-1)^k\binom{n+1}k\right]+(-1)^{n+1}\binom{n+1}{n+1}\ .$$
Unfortunately now the sum on the RHS does not fit your inductive hypothesis, so it is not going to work.
The easiest way is to expand $(1-1)^n$ as suggested by Lulu in comments.  However if you want an induction proof, the following will work.
I omit the basis step as you have done this yourself.
Suppose that
$$\sum_{k=0}^n(-1)^k\binom nk=0\ .$$
Using the Pascal's triangle recurrence,
$$\eqalign{
  \sum_{k=0}^{n+1}(-1)^k\binom{n+1}k
  &=\sum_{k=0}^{n+1}(-1)^k\left[\binom nk+\binom n{k-1}\right]\cr
  &=\sum_{k=0}^{n+1}(-1)^k\binom nk
    +\sum_{k=0}^{n+1}(-1)^k\binom n{k-1}\ .\cr}$$
In the first sum the $k=n+1$ term is zero, so drop it:
$$\sum_{k=0}^{n+1}(-1)^k\binom nk=\sum_{k=0}^n(-1)^k\binom nk\ .$$
In the second sum the $k=0$ term is zero, so drop it; then substitute $m=k-1$:
$$\sum_{k=0}^{n+1}(-1)^k\binom n{k-1}
  =\sum_{k=1}^{n+1}(-1)^k\binom n{k-1}
  =\sum_{m=0}^n(-1)^{m+1}\binom nm\ .$$
Thus
$$\eqalign{
  \sum_{k=0}^{n+1}(-1)^k\binom{n+1}k
  &=\sum_{k=0}^n(-1)^k\binom nk+\sum_{m=0}^n(-1)^{m+1}\binom nm\cr
  &=\sum_{k=0}^n(-1)^k\binom nk-\sum_{m=0}^n(-1)^m\binom nm\cr
  &=0-0\cr
  &=0\ .\cr}$$
A: There's an easier way without using induction.  Note that the expansion of $(1+(-1))^n$ is $\sum_{k=0}^{n}1^{n-k}(-1)^k\binom{n}{k}=\sum_{k=0}^{n}(-1)^k\binom{n}{k}$.
Then very simply $(1+(-1))^n=0$
A: Hints:
Find the expansion of $\left(1-x\right)^n$ and substitute $x=1$
Spoiler:

 $\quad\left(1-x\right)^n \\ = \sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix}1^{n-k}\left(-x\right)^k \\ =\sum_{k=0}^n \left(-1\right)^k x^k\begin{pmatrix} n\\k \end{pmatrix}\\ \text{Substitute both side by $x=1$ and we get}\\ \left(1-1\right)^n=0=\sum_{k=0}^n \left(-1\right)^k\begin{pmatrix} n\\ k \end{pmatrix}$

A: Hint for an algebraic proof: $(1-1)^n$.
Hint for a combinatorial proof: add or remove 1 to change parity.  For positive $n$, the identity can be rewritten as $$\sum_k \binom{n}{2k}= \sum_k \binom{n}{2k+1},$$ so we want to establish a bijection between even subsets and odd subsets of $\{1,\dots,n\}$. The bijection I hinted at is $$f(S)=\begin{cases}S \setminus \{1\}&\text{if $1\in S$},\\S \cup \{1\}&\text{otherwise}. \end{cases}$$
A: The last passage should actually be
$$
\begin{gathered}
  \sum\limits_{\left( {0\, \leqslant } \right)\,k\, \leqslant \,\left( {n + 1} \right)} {\left( \begin{gathered}
  n + 1 \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k} }  = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k} }  + \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k - 1 \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k} }  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k} }  + \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k + 1} }  = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k} }  - \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k} }  = 0 \hfill \\ 
\end{gathered} 
$$
Note that you can not take account of the
bounds in the summation because they are implicit
in the definition of the binomial.
A: The number of even sized subsets is the same as the number of odd subsets. 
To see this, note there’s a bijection because if 1 is in S, we can remove it. If 1 is not in S, we add it to S.
