How would I solve this mathematical induction proof? I am stuck after the first part of the induction. $$1 + 5 + 5^2 + \ldots + 5^n = \frac{5^{n+1}-1}{4}$$
Basis case $n= 0$: 
$1^0 = 1 \;\;\;\;\;\;\;\;\;\;\;\; \frac{5^{1+1}-1}{4}=1$
Assume true for $n=k$:
$$1 + 5 + 5^2 + \ldots + 5^k = \frac{5^{k+1}-1}{4}$$
Need to show for $n=k+1$:
$$1 + 5 + 5^2 + \ldots + 5^{k+1} = \frac{5^{k+1+1}-1}{4} = \frac{5^{k+2}-1}{4}$$
Induction proof
$$\frac{5^{k+1}-1}{4} + 5^{k+1}=\frac{5^{k+1}-1}{4} + \frac{4(5^{k+1})}{4}$$ This is where I am stuck and do not know if I am even right with this at all.

This is a geometric series, with common ratio = 5.
$$1 + q + q^2 + ... + q^n = (1 - q^{n+1})/(1-q)$$
Proof of this assumption by induction :
The case n=0 is trivial ($ 1 = 1 $)
Assume you have the previous result for n.
$$1+q+...+q^{n+1} = (1-q^{n+1})/(1-q) + q^{n+1}$$
$$1+...+q^{n+1} =(1-q^{n+1} + q^{n+1}\cdot (1-q) ) / (1-q)$$
If you develop, you get the Right hand side :
$$ (1 - q^{n+1} + q^{n+1} - q^{n+2})/(1-q)$$
Thus, 
$$(1-q^{n+2})/(1-q)$$
Which is the expected result for n+1
