A proof by induction could be done:
If $n=1, 2^n = 2^0 = 1$ and $1$ word of length $1$ has an even amount of $0$s, which is the word $1$.
Now imagine you have a sequence of length $k-1$. If it has an odd number of $0$s, then you can append a $0$ to the end of it. If it has an even number of $0$s, you can append a $1$ to it.
Assume then that it is true that, for words of length $k$, there are $2^{k-1}$ words with an even number of $0$s.
To those $2^{k-1}$ which have an even number of $0$s you add a $1$, making $2^{k-1}$ words of length $k+1$ that have an even amount of $0$s. To the other $2^{k-1}$ that have an odd number of $0$s, you add a $0$, making another $2^{k-1}$ words that have an even number of $0$s and length $k+1$. Therefore, you have a total of $2^{k-1} + 2^{k-1} = 2\cdot2^{k-1} = 2^k$ words, of length $k+1$, with an even number of $0$s, concluding your proof.