The last digit of 7^2026

The last digits of $7^1,7^2,7^3,7^4$ are $7,9,3,1$, and $7^4\equiv1\pmod{10}$, so the pattern repeats with period $4$.

Since $2026=4\cdot506+2$,
$$7^{2026}=\left(7^4\right)^{506}\cdot7^2\equiv49\equiv9\pmod{10}.$$

The last digit is $9$.