The identity $$\sum _{k=0}^n (-1)^k {n \choose k} = 0$$
For all $\mathbb{N}$ can be obtained from setting $a=-1$ and $b=1$ in the binomial thrm $(a+b)^n = \sum_{k=0}^n {n \choose k} a^k b ^{n-k}$. However, the former identity doesn't hold when $n=0$, as we have that the LHS thereof is $1$. However, the binomial theorem does hold when $n =0$, and because the binomial theorem implies this identity, it seems logically that this identity should hold when $n=0$, but it doesn't. Why is this?
There is no issue here. When you take $(x+y)^n=\sum_k \binom nk x^ky^{n-k}$ and set $x=1,y=-1$, you get $$ 0^n=\sum_{k=0}^n \binom nk (-1)^k $$ This is different from what you had; you wrote $0$ on the LHS, when it should have been $0^n$. Note that, in the context of combinatorics, and more generally in the context of integer-only arithmetic, the commonly adopted convention is $$ 0^0=1 $$ Therefore, $0$ is not an equivalent expression for $0^n$, because they differ when $n=0$.
The phenomenon you are observing is one of the very reasons why we define $0^0=1$; if we did not define $0^0$ this way, then the binomial theorem would not hold when $n=0$. For more reasons in favor of the $0^0=1$ convention for combinatorics and integer math, see Zero to the zero power – is $0^0=1$?.