7. The last digit of $\sum_{k=0}^{201}(10 k+7)^{k+1}$ is

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7.6.

It is easy to see that $\sum_{k=0}^{201}(10 k+7)^{k+1} \equiv \sum_{k=0}^{201} 7^{k+1}(\bmod 10)$.
Notice that, for any natural number $n$, the last digits of $7^{4 n+1}$, $7^{4 n+2}$, $7^{4 n+3}$, and $7^{4 n+4}$ are sequentially $7$, $9$, $3$, and $1$, and the last digit of their sum is 0.
$$
\text{Therefore, } \sum_{k=0}^{201}(10 k+7)^{k+1} \equiv 7^{1}+7^{2} \equiv 6(\bmod 10) \text{. }
$$