Suppose $y=0$ in
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$$
Then we get $\binom{n}{0}x^{n-0}0^0$ as the first term of the sum. We treat this as a 1, normally, though $0^0$ isn't well-defined. What am I missing? Or should $x$ and $y$ be required to be nonzero?
On
It is often good to define $0^0=1$, but there are some caveats. Namely, $$ \lim_{x\to0}\lim_{y\to0}x^y = 1 \neq 0 = \lim_{y\to0}\lim_{x\to0}x^y. $$ In a combinatorial setting this issue does not arise.
If you want a reason for why $0^0=1$ "combinatorically", consider the following. For $n,m$ positive integers there are $n^m$ different functions from a set of $m$ elements to a set of $n$ elements. What if $n=0$ or $m=0$? If $m=0$ and $n\geq0$, there is only one function (the empty function). If $m>0$ and $n=0$, there is no function, since there is no possible value to assign to your $m$ points.
On
In the sum $$(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k={n\choose0}x^ny^0+\sum_{k=1}^n \binom{n}{k}x^{n-k}y^k$$ the first term (which has been extracted above) ALWAYS has $k=0$; $k$ is not a variable (notice there is no $k$ on the LHS). The sum on the RHS has no restrictions on $x,y$, and the first term has the restriction $y\neq0$. But we take $$\lim_{y\to0}{n\choose0}x^n y^0={n\choose0}x^n$$ From this we can define $0^0=1$ so we do not have to take a limit everytime, and receive the same result. By the way it is approached in this context, it is the most natural way to define $0^0$. This is why, for example, we do not define $0^0=\lim_{y\to0}0^y=0$ here. In other contexts however, the just-mentioned definition may be more convenient or natural.
The reason $0^0$ is not defined is that there is no consistent way of defining it - the limit of $y^0$ as $y \to 0$ is clearly $1$, while the limit of $0^y$ as $y \to 0^+$ is clearly zero. But it can be formally defined for any particular application.
Here if you let $y$ approach $0$ with $n$ fixed you discover that the limit you need to make the expression continuous in $y$ is $0^0=1$ - so you can formally state that this is the value you are taking for this application.