What is the least positive integer $m$ such that $m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ is a perfect square?
$\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001$
Consider 2 --
there are odd number of 2's in $2!\cdot3!\cdot4!\cdot5!...16!$  (We're not counting 3 2's in 8, 2 3's in 9, etc).
There are even number of 3's in 
...etc,

So, we can reduce our original expression to
\begin{align*} m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ &\equiv m \cdot 2 \cdot 3 \cdot (2 \cdot 2) \cdot 5  \cdot (2 \cdot 3)  \cdot 7  \cdot (2  \cdot 2 \cdot 2)\\ &\equiv m  \cdot 2 \cdot 5  \cdot 7\\ m &= 2 \cdot 5 \cdot 7 = 70 \end{align*}
~Technodoggo ~minor edits by lucaswujc