If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$
$7^1$ has a unit digit of $7$.
$7^2$ has a unit digit of $9$.
$7^3$ has a unit digit of $3$.
$7^4$ has a unit digit of $1$.
$7^5$ has a unit digit of $7$.
Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$.  It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit.  Since $753$ leaves a remainder of $1$ after being divided by $4$, the units digit of $2137^{753}$ is $7$, which is answer choice $\boxed{\textbf{(D)}}$.