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 $p|\frac{p^{11}}{d^2}$.

Proof by example is satisfactory only if your examples exhaust the possibilities.
Assuming $d$ is positive, you could have only one of $d=1,p,p^2,p^3,p^4$, or $p^5$. If you can show that this is the list of all possibilities, you can 
verify the claim by treating these 6 cases one by one. 
You can simplify further by combining them to a single case $d=p^k$ with $0\le k\le 5$.
