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

In the following $p$ is a prime and $a,c,d,k,n \in \mathbb{N}$. Note the following:


*

*If $a^2=p^k$ then $k$ is even. This follow from unique factorization theorem.

*If $a|p^{n}$ then $a=p^k$ for some $k \leq n$. Write $p^k=c \cdot a$ and use again unique factorization theorem.


Since $d^2|p^{11}$ from 1 and 2 we conclude that $d^2=p^k$ for an even $k$ with $k \leq 11$. Because $k$ is even $ \Rightarrow k < 11$. 
Therefore $\frac{p^{11}}{d^2}=p^{11-k}$. Since $11-k>0 \Rightarrow 11-k\geq 1 \Rightarrow p|\frac{p^{11}}{d^2}$. 
