Proving with Big O Notations Is there a way I can prove that $O(3^{2n})$ does NOT equal $10^n$? How would that be done? Also, is it okay to simplify $O(3^{2n})$ to $O(9^n)$ to do so?

It seems that you want to show that $10^n\notin O(3^{2n})$.
To prove that $10^n$ is not $O(3^{2n})$ it is enough to show that for any $M$ there is $n$ such that $10^n > M\cdot3^{2n}$, in particular
\begin{align}
10^n &> M\cdot 9^n\\
\frac{10^n}{9^n} &> M \\
n &> \log_{\frac{10}{9}} M
\end{align}
so $n = \left\lfloor\frac{\log M}{\log\frac{10}{9}}\right\rfloor + 1$ suffices.
I hope this helps $\ddot\smile$
