Proof that $\sum_{k=0}^{n} {n\choose k}=2^n$ and implications Here is a case that I found very interesting:

Prove that $$\sum_{k=0}^{n} {n\choose k} = 2^n$$
where $n,k\in\mathbb Z^+$

So the proof for this statement is given by what I consider to be one of the most satisfying ones out there:
Proof
Consider $(x+1)^n$, by the binomial theorem we have: $$(x+1)^n=\sum_{k=0}^{n}\binom{n}{k} x^{n-k}1^k$$ $$(x+1)^n={n\choose0}x^{n-0}1^0+{n\choose1}x^{n-1}1^1+{n\choose2}x^{n-2}1^2+\ldots+{n\choose n}x^{n-n}1^n$$ given that $1^k=1$ holds  $\forall k \in \mathbb R$ $$(x+1)^n={n\choose 0}x^n + {n\choose 1}x^{n-1} + {n\choose 2}x^{n-2}+\ldots+{n\choose n}$$ now let $x=1$, hence $(x+1)^n = 2^n$ and $$2^n = {n\choose 0} + {n\choose 1} + {n\choose 2}+\ldots+{n\choose n}$$ $$\implies \sum_{k=0}^{n} {n\choose k}=2^n$$
The question that I've got is, since I found this case as intriguing as I did, I was wondering what implications this idea has (if any, of course) and whether or not there are any other interesting proofs that sprung from this one, or are similar to this one. Thank you!
 A: Here a proof through induction using the recursive relation:
Base case: If $n=0$, then $\sum_{k=0}^n \binom{n}{k} = \binom{0}{0} = 1 = 2^0.$
Induction step: Suppose that the sum $\sum_{k=0}^n\binom{n}{k} = 2^n$, then
$$\sum_{k=0}^{n+1} \binom{n+1}{k} =2 + \sum_{k=1}^n\binom{n+1}{k} = 2 + \sum_{k=1}^n\binom{n}{k-1} + \sum_{k=1}^n\binom{n}{k}$$
$$= 2 + \sum_{k=0}^{n-1}\binom{n}{k} + \sum_{k=1}^n\binom{n}{k} = 2 +  2\sum_{k=0}^{n}\binom{n}{k} - \binom{n}{0} -\binom{n}{n}$$
$$=2 \sum_{k=0}^{n}\binom{n}{k} = 2^{n+1}.$$
A: This identity is in fact a weakened form of the Binomial theorem,
$$\sum_{k=0}^n\binom nk a^kb^{n-k}=(a+b)^n$$ where you just substitute $(1,1)$ for $(a,b)$.
$$\sum_{k=0}^n\binom nk=(1+1)^n.$$
Similarly,
$$\sum_{k=0}^n\binom nk(-1)^k=0.$$

The result can also be easily obtained from a well-known property of Pascal's triangle: every element is the sum of the two elements above it (Pascal's rule).
Then the sum of the elements of a row is twice the sum of the elements of the row above.
A: Here is another proof by double counting:
How can we create a committee of any size from a pool of $n$ candidates? We can separate this into the cases where the committee is of size $k$, of which there are $\binom{n}{k}$ such committees. Summing over all values of $k$ accounts for all sizes of the committee to obtain $\sum_{k=0}^n\binom{n}{k}$ total committees. Another answer to the question is to look at each candidate and decide whether or not they are on the committee. There are two options per person: on the committee or not, so there are $2^n$ ways to choose this committee of arbitrary size. Hence, $\sum_{k=0}^n\binom{n}{k}=2^n$.
I don't know of any proofs coming out of the one you give, but we can use the binomial theorem to develop more identities similar to what you have. For instance, putting $x=1,y=-1$ into the binomial theorem formula, we have $0=\sum_{k=0}^n\binom{n}{k}(-1)^k$.
