AHSME 1951 Problem 24

$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is: $\textbf{(A)}\ 2^{n+1}-\frac{1}{8} \textbf{(B)}\ -2^{n+1} \textbf{(C)}\ 1-2^{n} \textbf{(D)}\ \frac{7}{8} \textbf{(E)}\ \frac{7}{4}$