last 4 digits of $7^{65}$

Question

As the name implies, what are the last 4 digits of $7^{65}$

I can't find anything that does not involve bashing, I tried $(8-1)^{65}$, or maybe $7^{65}$ mod 625 and 16,

Answer 1

$7 (7^2)^{32} = 7\color{#c00}{(50\!-\!1)^{32}} = 7 \overbrace{\left[1 - 32\cdot 50 + \dfrac{32\cdot 31}2\ 50^2 +10^4(\cdots)\right]}^{\rm\large appply\ \color{#c00}{Binomial\ Theorem}} \equiv 8807\pmod{\!10^4} $

Answer 2

If you calculate the first powers of 7, you will see quickly see the pattern that develops - for example, the last digit is going 7, 9, 3, 1, 7, 9, 3, 1, ... - I can calculate that in my head.
Simply try the first ten or fifteen powers and write them down, and you'll see the pattern.