<s>[INST] 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.
 [/INST] Hint:
$$\color{red}{4^3+\cdots 4^n}+4^{n+1}=\color{red}{\frac{4(4^n-16)}{3}}+4\cdot4^{n}=\left(\frac43+4\right)4^n-4\cdot\frac {16}3$$
</s>