When $p = \sum\limits_{k=1}^{6} k \text{ ln }{k}$, the number $e^p$ is an integer.  What is the largest power of $2$ that is a factor of $e^p$?
$\textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}\ 2^{18}\qquad\textbf{(E)}\ 2^{20}$
Let's write out the sum.  Our sum is equal to 
\[1 \ln{1} + 2 \ln{2} + 3 \ln{3} + 4 \ln{4} + 5 \ln{5} + 6 \ln{6} =\]
\[\ln{1^1} + \ln{2^2} + \ln{3^3} + \ln {4^4} + \ln{5^5} + \ln {6^6} =\]
\[\ln{(1^1\times2^2\times3^3\times4^4\times5^5\times6^6)}\]
Raising $e$ to the power of this quantity eliminates the natural logarithm, which leaves us with 
\[e^p = 1^1\times2^2\times3^3\times4^4\times5^5\times6^6\]
This product has $2$ powers of $2$ in the $2^2$ factor, $4*2=8$ powers of $2$ in the $4^4$ factor, and $6$ powers of $2$ in the $6^6$ factor.
This adds up to $2+8+6=16$ powers of two which divide into our quantity, so our answer is $\boxed{\textbf{(C)}\ 2^{16}}$