Define a sequence of real numbers $a_1$, $a_2$, $a_3$, $\dots$ by $a_1 = 1$ and $a_{n + 1}^3 = 99a_n^3$ for all $n \geq 1$. Then $a_{100}$ equals
$\textbf{(A)}\ 33^{33} \qquad  \textbf{(B)}\ 33^{99} \qquad  \textbf{(C)}\ 99^{33} \qquad  \textbf{(D)}\ 99^{99} \qquad  \textbf{(E)}\ \text{none of these}$
We rearrange to get $\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}$. Thus we get $\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}$, $\dfrac{a_{n}}{a_{n-1}} = \sqrt[3]{99}$, and so on. Multiplying them all gives $\dfrac{a_{n+1}}{a_1} = (\sqrt[3]{99})^{n}$. Plugging in $n = 99$ and $a_1 = 1$, $a_{100} = (\sqrt[3]{99})^{99} = 99^{33}$, so the answer is $\textbf{(C)}$.