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?
 A: 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 $0^0=1$.
A: Defining $0^0=1$ is required if we want that the binomial theorem be true for $n=0$:
$$1=(1+0)^0=\binom{0}{0}1^0 0^0=\cdots$$
A: Another "proof" which I think is pretty neat:
$$
\begin{align}
(1-1)^n &= \sum_{k=0}^n\binom{n}{0}\times(-1)^k1^{n-k}\\
& = \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\dots\binom{n}{n}
\end{align}
$$
This corresponds to the alternating series of the summing of a row in Pascal's triangle:
             
             
             
             
      
Which corresponds $0^n$, where $n$ is the $(n+1)$th row, and will almost always be 0. Now you can easily check this being true for any row, except for the first one, which corresponds to $0^0$. Now what is the alternating series with a single $1$?
A: As seen in other answers, the binomial theorem, in its usual formulation, implies $0^0=1$  (why this happens will be explained in the last paragraph).  This leaves 4 choices:
(a) Accept $0^0=1$ as fact.
(b) Reformulate the binomial theorem so that it no longer implies $0^0=1$.
(c) Abandon the principle that consequences of widely used theorems must be accepted as fact.
(d) Acknowledge that math is not consistent.
I find (d) completely unacceptable.
Option (c) is not different.
Option (b) does not solve the problem either, because even if we restrict the binomial theorem
so that it no longer implies $0^0=1$, the more general form would still be widely used, and even
if we managed to address that, there would still be other theorems that imply $0^0=1$.
That leaves option (a).
Why does the binomial theorem imply $0^0=1$? Even if you don't want $0^0$ to be defined, it is very easy to accidentally prove a theorem that implies $0^0=1$. How? Say $P_1$ and $P_2$ are products, but $P_1$ is an empty product (e.g. $x^0$), then $P_1 P_2$ is almost universally viewed as $P_2$. But this "innocent" step means that $P_1$ is interpreted as $1$. This way $x^0$ ends up being viewed as $1$ for all $x$, including $0$. Thus, a theorem is proved that unintentionally implies $0^0=1$. In summary: unless one splits the proof into more cases (and why would you want to do that?) to actively avoid empty products, it's easy to end up proving something that implies $0^0=1$ even if you don't want to.
