Let N = $69^{5} + 5\cdot69^{4} + 10\cdot69^{3} + 10\cdot69^{2} + 5\cdot69 + 1$. How many positive integers are factors of $N$?
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 69\qquad \textbf{(D)}\ 125\qquad \textbf{(E)}\ 216$
Let $69=a$. Therefore, the equation becomes $a^5+5a^4+10a^3+10a^2+5a+1$. From Pascal's Triangle, we know this equation is equal to $(a+1)^5$. Simplifying, we have the desired sum is equal to $70^5$ which can be prime factorized as $2^5\cdot5^5\cdot7^5$. Finally, we can count the number of factors of this number.
$6\cdot6\cdot6=\fbox{(E) 216}$.