Binomal theorem show that $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots=2^{n-1}$ I'm having some trouble with this question

Show that
  $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots=2^{n-1}$$

Attempt:
Expanding $(1+1)^n=2^n$
$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\dots+\binom{n}{n}=2^n$$
Expanding $(1-1)^n=0$
$$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}+\dots+(-1)^n\binom{n}{n}=0$$
And I'm stuck here. The solution says "Adding these two equalities yields the required one".
I can see how the terms will cancel out,however I don't see how $2^n+0 = 2^{n-1}$

Also you can subtract equation 2) from 1) to get second part.
