If $d^2|p^{11}$ where $p$ is a prime, explain why $p|\frac{p^{11}}{d^2}$. If $d^2|p^{11}$  where $p$ is a prime, explain why $p|\frac{p^{11}}{d^2}$.
I'm not sure how to prove this by way other than examples.  I only tried a few examples, and from what I could tell $d=p^2$.  Is that always the case?
Say $p=3$ and $d=9$.  So, $9^2|3^{11}$ because $\frac{3^{11}}{9^2}=2187$.  Therefore, $3|\frac{3^{11}}{9^2}$  because $\frac{2187}{3}=729$.  Is proof by example satisfactory?  
I know now that "proof by example" is only satisfactory if it "knocks out" every possibility.
The proof I am trying to form (thanks to the answers below):
Any factor of $p^{11}$ must be of the form $p^{k}$ for some $k$.
If the factor has to be a square, then it must then be of the form $p^{2k}$, because it must be an even power.
Now, we can show that $\rm\:p^{11}\! = c\,d^2\Rightarrow\:p\:|\:c\ (= \frac{p^{11}}{d^2})\:$ for some integer $c$.
I obviously see how it was achieved that $c=\frac{p^{11}}{d^2}$, but I don't see how what has been said shows that $p|\frac{p^{11}}{d^2}$.

Hint $\ $ It suffices to show $\rm\:p^{11}\! = c\,d^2\Rightarrow\:p\:|\:c\ (= p^{11}\!/d^2).\:$  We do so by comparing the parity of the exponents of $\rm\:p\:$ on both sides of the first equation. Let $\rm\:\nu(n) = $ the exponent of $\rm\,p\,$ in the unique prime factorization of $\rm\,n.\:$ By uniqueness $\rm\:\color{#C00}{\nu(m\,n) = \nu(m)+\nu(n)}\:$ for all integers $\rm\:m,n\ne 0.\:$ Thus
$$\rm \color{#C00}{applying\,\ \nu}\ \ to\ \ p^{11}\! =\, c\,d^2\ \Rightarrow\ 11 = \nu(c) + 2\, \nu(d)$$
Therefore $\rm\:\nu(c)\:$ is odd, hence $\rm\:\nu(c) \ge 1,\:$ i.e. $\rm\:p\mid c.\ \ $ QED
