Evaluating Summation of $5^{-n}$ from $n=4$ to infinity The answer is $\frac1{500}$ but I don't understand why that is so. 
I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}= \frac54$.

The problem you have is that you do not know why the formula works to begin with. If you did the situation would be clear. Here's the thing:
$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r} \; \;\;\;\;\; |r|<1.$$
Let $r=\frac{1}{5}$, then you really have the following situation:
$$\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
Let's call this infinite sum $S$ and proceed as follows,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
then 
$$ \left(\frac{1}{5}\right)S=\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+\left(\frac{1}{5}\right)^6....\;\;\;\;$$
Subtract the second from the first,
$$S-\left(\frac{1}{5}\right)S=1$$
$$S(1-\left(\frac{1}{5}\right))=1$$
$$S=\frac{1}{1-\left(\frac{1}{5}\right)}$$
$$S=\frac{5}{4}.$$
Now recall what $S$ was and realize,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}$$
But, you want the powers to start at $n=4$, so subtract the first 4 powers(0,1,2,3) to get,
$$S-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)$$
which means 
$$\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)=\frac{1}{500}.$$
