Is my $0^0 = 1$ proof correct? Is the following proof that $0^0 = 1$ correct?
Let $k = 0^0$ (assuming it has a meaningful value)
Observe that $k = 0^0 = 0^{n0} = (0^0)^n = k^n$, for all integers $n$.
$\implies k^n - k = 0$, for all $n \in \mathbb Z$.
$\implies k^n - k = 0$ and $k - k^{-n} = 0$, for all $n \in \mathbb N$.
$\implies k[k^{n-1} - 1] = 0$ and $\displaystyle\frac1{k^n}[k^{n+1} - 1] = 0$
$\implies [k = 0$ or $k =$ any $(n-1)$th root of unity$]$ and $[k =$ any $(n+1)$th root of unity$]$
All requirements are satisfied $\iff k = 1$ for any fixed $n$.
Therefore, $k = 0^0 = 1\quad\blacksquare$

Just as an aside - a more 'proofy' way (if that even makes sense!) to show this is to calculate the (right) limit as $x \rightarrow 0$ of the function $x^x$ and to take this as the definition of $0^0$. 
To do this simply note that
$$x^x = e^{x\log x}. $$
Then standard limit rules give
$$0^0 = \lim_{x \rightarrow 0^+} e^{x\log x}= e^0 = 1.$$
