The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$, where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad  \textbf{(B)}\ 4n \qquad  \textbf{(C)}\ 2+2n \qquad  \textbf{(D)}\ 4n^2 \qquad  \textbf{(E)}\ n^2+n+2$
We see that the result of this expression will always be in the form $(100\text{ some number of zeros }001)^2.$ Multiplying these together yields: \[110\text{ some number of zeros }011.\] This works because of the way they are multiplied. Therefore, the answer is $\boxed{(A) 4}$.