Disproving $ 0^0 $ by binomial theorem 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.
