Let $S(n)$ denote the sum of digits of a non-negative integer n. Prove that $S(1981^n) \geq 19$
I tried to use induction. I know that $S(pq) \leq S(p)S(q)$ so $S(1981^{n+1}) \leq 19S(1981^n)$ but it leads to nowhere.
My second thought was to use $S(n) = n - 9\sum_{i\ge 1} \left\lfloor\frac n{10^i}\right\rfloor$ formula but I find it's useless.
Any help is appreciated.
The formula for $S(n)$ along with the fact that $1981 \equiv 1 \pmod{9}$ gives that for all $n\in \mathbb{N}$ $$S(1981^n)\equiv 1 \pmod{9}$$ So we just need to show that $S(1981^n)$ can never be equal to $10$. Suppose that for some $n$ we have $S(1981^n)=10$. Then $$1981^n - 1 = \overline{d_kd_{k-1}\dots d_10}$$ with $d_k+d_{k-1}+\dots+d_1 \equiv 0 \pmod{9}$. The first equality comes from the fact that $1981^n - 1$ is divisible by $10$ (since $1981\equiv 1 \pmod{10}$) while the second equivalence follows form the divisibility rule for $9$. This gives that $$10 = S(1981^n) = \sum_{i=1}^k d_i + 1$$ so $$d_k+d_{k-1}+\dots+d_1=9 \tag{1}$$ Doing the exact same thing modulo $11$ (since $1981\equiv 1\pmod{11}$) we get $$d_k-d_{k-1}+\dots+(-1)^{k-1}d_1\equiv 0 \pmod{11}$$ But from $(1)$ the only way this can happen is if $$d_k-d_{k-1}+\dots+(-1)^{k-1}d_1=0 \tag{2}$$
Adding $(1)$ and $(2)$ gives a contradiction since the LHS will be even while the RHS will be $9$.