Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.

Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$

I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
My confusion is on the induction step where:
$4^3+4^4+4^5+⋅⋅⋅+4^n+4^{(n+1)} = 4(4^{(n+1)}-16)/3$.
I don't know what to do next.
 A: Hint:
$$\color{red}{4^3+\cdots 4^n}+4^{n+1}=\color{red}{\frac{4(4^n-16)}{3}}+4\cdot4^{n}=\left(\frac43+4\right)4^n-4\cdot\frac {16}3$$
A: By the induction hypothesis 
$$
4^3+4^4+\ldots+4^n=\frac{4(4^n-16)}{3}=\frac{4^{n+1}-64}{3}
$$
so 
$$
4^3+4^4+\ldots+4^n+4^{n+1}=\frac{4^{n+1}-64}{3}+4^{n+1}=\frac{4^{n+1}-64+3\cdot4^{n+1}}{3}\\
=\frac{4\cdot 4^{n+1}-64}{3}=\frac{4^{n+2}-64}{3}=\frac{4(4^{n+1}-16)}{3}
$$
QED
A: When doing a proof by induction, I suggest you say so up front.
Label the base case and the inductive hypothesis.  Here is your template.
Prove:
$4^3 + 4^4 + 4^5 \cdots 4^n = 4(4^n - 16)/3$
Proof by induction.
Base case:
$4^3 = 4(4^3-16)/6$
Inductive hypothesis:
Suppose, $4^3 + 4^4 + 4^5 \cdots 4^n = 4(4^n - 16)/3$
We must show that,
$4^3 + 4^4 + 4^5 \cdots 4^{n+1} = 4(4^{n+1} - 16)/3$
based on the inductive hypothesis.
$4^3 + 4^4 + 4^5 \cdots 4^{n+1}\\
4^3 + 4^4 + 4^5 \cdots 4^n + 4^{n+1}$
$\frac {4(4^n-16)}{3} + 4^{n+1}$ from the inductive hypothesis.
$\frac {4(4^n-16) + 3(4^{n+1})}{3}\\
\frac {4(4^n-16 + 3\cdot4^{n})}{3}\\
\frac {4(4^{n+1}-16)}{3}$
QED
