For fun, is something like this true?
Let \begin{equation} \nonumber \begin{split} k &= 0^0 = (a-a)^{a-a} = \frac{(a-a)^{a}}{(a-a)^{a}} \\ &= \frac{\binom{a}{0}a^a(-a)^0 + \binom{a}{1}a^{a-1}(-a)^1 + ... + \binom{a}{a-1}a^{1}(-a)^{a-1} + \binom{a}{a}a^{0}(-a)^{a}}{\binom{a}{0}a^a(-a)^0 + \binom{a}{1}a^{a-1}(-a)^1 + ... + \binom{a}{a-1}a^{1}(-a)^{a-1} + \binom{a}{a}a^{0}(-a)^{a}}. \\ \end{split} \end{equation}
Furthermore, let $a$ be any odd number, then
\begin{equation} \nonumber \begin{split} k &= \frac{\binom{a}{0}a^a - \binom{a}{1}a^{a} + ... + \binom{a}{a-1}(a)^{a} - \binom{a}{a}(a)^{a}} {\binom{a}{0}a^a - \binom{a}{1}a^{a} + ... + \binom{a}{a-1}(a)^{a} - \binom{a}{a}(a)^{a}} \\ \end{split} \end{equation}
since $a$ is odd there is an even number of terms in the polynomials, thus
\begin{equation} \nonumber \begin{split} k = \frac{0}{0}. \\ \end{split} \end{equation}
You are not disproving anything. $0^0$ is a mathematical expression with no agreed-upon value. The most common possibilities are $k=1$ or leaving the expression undefined, with justifications existing for each, depending on context.