. Determine the units digit of $1789^{1789}$.

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. On a: $1789 \equiv 9(\bmod 10) \Rightarrow 1789^{1789} \equiv 9^{1789}(\bmod 10)$.

Let's look at the powers of 9 modulo 10. We have: $9^{0} \equiv 1(\bmod 10), 9^{1} \equiv 9$ $(\bmod 10), 9^{2} \equiv 1(\bmod 10), 9^{3} \equiv 9(\bmod 10)$, etc. Then for any natural number $k$, $9^{2 k} \equiv 1(\bmod 10)$ (indeed, $9^{2 k}=81^{k}$ and $81 \equiv 1(\bmod 10)$) and $9^{2 k+1} \equiv 9$ $(\bmod 10)$ (indeed, $9^{2 k+1}=9^{2 k} \times 9$ and $9^{2 k} \equiv 1(\bmod 10)$). Since 1789 is odd, we deduce that: $9^{1789} \equiv 9(\bmod 10)$. Thus, the units digit of $1789^{1789}$ is 9.