Proving $\sum_{k=0}^{n} {n \choose k} = 2^n$ with Newton's Binomial Theorem

Question

I'm having a hard time proving this theorem from a textbook.

Theorem

For any integer $n \ge 0$, we have $$\sum_{k=0}^{n} {n \choose k} = 2^n$$

Proof

Take $x = y = 1$ in Newton's Binomial Theorem

My work so far

$$ x = y = 1 \\ \sum_{k=0}^n{n \choose k} 1^{n-k}1^k \\ = \sum_{k=0}^n{n \choose k} 1^n $$

Then I tried using concrete examples $$ k=0,\ \binom{n}{0}1^n \qquad \qquad \binom{n}{0} = \frac{n!}{0!(n-0)} = \frac{n!}{n!}=1\\ k=1,\ \binom{n}{1}1^n \qquad \qquad \binom{n}{1} = \frac{n!}{1!(n-1)} = \frac{n!}{(n-1)!}\\ k=2,\ \binom{n}{2}1^n \qquad \qquad \binom{n}{2} = \frac{n!}{2!(n-2)} \\ \vdots \\ k=n,\ \binom{n}{n}1^n \qquad \qquad \binom{n}{n} = \frac{n!}{n!(n-n)} =1\\ =1^n+\cdots+1^n $$

How to prove this?

Answer 1

Note that $1$ to any power is $1$, that is, $1^{n-k}=1=1^{n}$ and so $$2^{n}=\left(1+1\right)^{n}=\sum_{k=0}^{n}{n \choose k}1^{n-k}\cdot1^{k}=\sum_{k=0}^{n}{n \choose k}.\text{}$$

Answer 2

Hint. The binomial theorem is an equality, not just an expression. You have written down $$\sum_{k=0}^n{n \choose k} x^{n-k}y^k\tag{*}$$ and substituted $x=y=1$. That's good so far, but the binomial theorem says that $(*)$ is equal to something else. You need to remember (or look up) what this "something else" is, and substitute $x=y=1$ into that as well.