combinatorics - Prove that $0^0 = 1$ using binomial theorem

I have read in Donald Knuth's book that $0^0 = 1$. And it has been said to come from the basic formula of $(x+y)^r$. Can anyone prove how it comes?

Answer

Consider

$$
(x+0)^n=\color{#00A000}{\binom{n}{0}x^n0^0}+\color{#C00000}{\binom{n}{1}x^{n-1}0^1+\dots+\binom{n}{n}x^00^n}
$$
Since all the red terms are $0$, for the left side, $x^n$, to equal the right side, $x^n0^0$, we need $0^0=1$.

Furthermore, we have
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}=(1-1)^n
$$
Note that this holds for $n=0$ only if $0^0=1$.