Inspired by Chinmay Sangawadekar

Like Zee, I will try to explain this without (explicitly) using congruences. To save us a little time, we will let $a=3125$ throughout. We can ignore the first digit 4 in 43125 since it will not affect the last four digits of the power.

We will show that $a^{501}-a$ is divisible by 10000; that implies that the last four digits of $a^{501}$ are $a=\boxed{3125}$ . Since $10000=5^4\times2^4$ , it suffices to show that $a^{501}-a$ is divisible by both $5^4=625$ and $2^4=16$ .

We can factor $b=a^{501}-a=a(a^{125}-1)(a^{125}+1)(a^{250}+1)$ . Now $b$ is divisible by $5^4$ since $a=3125=5^5$ . Also, $b$ is divisible by 16 since three of the factors are even and one (either $a^{125}+1$ or $a^{125}-1$ ) is divisible by 4.

An elementary solution (without writing out congruences):

If we want to determine the last four digits of a(n integral) power (of an integer), we can do this in multiple steps, taking only the last for digits of our results into account at the next step. (This is easy to prove, by examining the last four digits (of m×n) of the (10000+m)×(10000+n) product in the m=n case, by induction.)

Now, $3125^4$ = 95,367,640,625 , therefore we can calculate the last four digits of $3125^{20 }$ as the last four digits of $(3125^4)^5$ , which is the same as the last four digits of $625^5$ . We will get 0625 again, since $3125^4=625^5=5^{20}$ .

Putting it to the 5th power twice more, we will find, that that the last four digits of both $3125^{100}$ and $3125^{500}$ are 0625.

The last step is to determine the the last four digits of $3125^{501}$ , which is the same as the last four digits of 625×3125=1,953,125. Therefore, our answer is $\boxed {3125}$ .