<s>[INST] How do you prove $37^{100} - 37^{20}$ is a multiple of $10$ using induction? I've tried making it $37^{5n} - 37^n$ is a multiple of $10.$ Then I made the base case be $n = 0,$ so $1 - 1 = 0$ which is a multiple of $10.$ I assumed $37^{5n} - 37^n$ is a multiple of $10$ for all $n \geqslant0.$ The inductive step is really hard however. I did $37^{5(n + 1)} - 37^{n+1}$ which I simplified to $37^{5n + 5} - 37^{n+1}.$ I tried factoring and made it $37^5\cdot37^{5n} - 37\cdot(37^n).$ I don't know where to go from here though. 
 [/INST] I know OP requested induction, but I would like to post this anyway on the off-chance that someone finds it interesting:
This is the same as proving that $10 {\large\mid} 37^{80}-1$, since $37^{100}-37^{20} = 37^{20}(37^{80}-1)$ and $37^{20}$ has only powers of $37$ as divisors since $37$ is prime.
Clearly $2{\large\mid}37^{80}-1$ since $37^{80}-1$ is even, so we must only show that $5{\large\mid}37^{80}-1$. Observe that $37 \equiv 2 \pmod{5}$, therefore $37^{80} \equiv 2^{80} \pmod{5}$. Thus we have $2^{80} \equiv (2^{10})^8 \equiv (1024)^8 \equiv (4)^8 \equiv (4^2)^4 \equiv 1^4 \equiv 1 \pmod{5}$.
Thus $37^{80} \equiv 1 \pmod{5}$, and so $37^{80} - 1 \equiv 0 \pmod{5}$, as desired.
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