Example 7 (1) Prove: For any natural number $a, a^{4 n+k}$ has the same units digit as $a^{k}$ $(n$, $k \in \mathrm{N}$ );
(2) Find the units digit of $2^{1997}+3^{1997}+7^{1997}+9^{1997}$.
Proof: (1) It needs to be proven that $a^{4 n+k}-a^{k} \equiv 0(\bmod 10)$
Since $a^{4 n+k}-a^{k}=a^{k}\left(a^{4 n}-1\right) \equiv 0\left(\bmod a\left(a^{4}-1\right)\right)$, it is only necessary to prove that $a\left(a^{4}-1\right) \equiv 0(\bmod 10)$.
And $a\left(a^{4}-1\right)=(a-2)(a-1) a(a+1)(a+2)+5(a-1) a(a+1)$. Therefore, (2) holds, and thus (1) holds.
(2) Using (1), we have
$$
2^{1997}+3^{1997}+7^{1997}+9^{1997} \equiv 2+3+7+9 \equiv 1(\bmod 10),
$$

which means the units digit of $2^{1997}+3^{1997}+7^{1997}+9^{1997}$ is 1.