Proving a special case of the binomial theorem: $\sum^{k}_{m=0}\binom{k}{m} = 2^k$ I want to know if I can get some help with this proof. I tried, but I just cannot seem to get $2^{k}$. It states that,

For $k \in \mathbb{Z}_{\ge 0}$, $$\sum^{k}_{m=0}\binom{k}{m} = 2^k$$

Thank You.
 A: The right-hand side is the number of subsets of a set with $k$ elements.  The left hand side gives a sum of the number of subsets of a set with $k$ elements having $0$ elements, then $1$ element, then $2$ elements, etc., up to the subsets with $k$ elements.
Leaving that aside, this follows from applying the binomial theorem in a way that might seem like a trick.  Here's another example to motivate this.  If $k\geq 1$, then
$${k \choose 0}-{k \choose 1}+{k \choose 2}\mp\cdots +(-1)^k{k \choose k}=0.$$
Proof Sketch: Expand $(1-1)^k$ using the binomial theorem.
You could use induction and the factorial formula for the binomial coefficients, but I do not recommend doing so.  However, using induction and Pascal's identity ${k\choose {m-1}}+{k\choose m}={{k+1}\choose m}$ would work well.  Consider moving down a row in Pascal's triangle to motivate the proof.  Each entry from row $k$ is added twice to obtain the entries in row $k+1$, so the sum of the entries doubles.
A: Take the binomial expansion of $(1+x)^k$.
\begin{equation}
(1+x)^k = \sum_{m=0}^k \binom{k}{m} x^{k-m}
\end{equation}
The above expansion holds for all values of $x$ and for all $k \geq 0$. Substitute $x=1$ and you have the result.
A: You can use induction.
$$\binom{k}{m}=\binom{k-1}{m-1}+\binom{k-1}{m}$$ 
is true for $k\in\mathbb{Z}_{>0}$
and $m\in\mathbb{Z}$. 
Here: $$\binom{k}{m}:=0$$ if $m\notin\left\{ 0,\ldots,k\right\} $.
So: 
$$\sum_{m\in\mathbb{Z}}\binom{k}{m}=\sum_{m\in\mathbb{Z}}\binom{k-1}{m-1}+\sum_{m\in\mathbb{Z}}\binom{k-1}{m}=2^{k-1}+2^{k-1}=2^{k}$$
A: The LHS of the equation is counting the number of subsets of a set of $k$ elements. The RHS of the equation is a well-known formula for that.
