<s>[INST] How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,

Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?

At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how to proceed further. Secondly I thought of using the fact $(a+b+\cdots) \pmod {17} = (r_a+r_b\dots) \pmod {17}$ but it is getting more messier.
Please explain in detail. And also mention the formula being used.
 [/INST] HINT:
Observe that for any non-negative integer $a,$ $$1+4+4^2+4^3=(1+4)(1+4^2)\equiv0\pmod{17}$$
$$\implies\sum_{a=m}^n4^{4a}(1+4+4^2+4^3)\equiv0\pmod{17}$$
Here $m=0,n=9$
</s>