ILL 1976 Problem 19

For a positive integer $n$ , let $6^{(n)}$ be the natural number whose decimal representation consists of $n$ digits $6$ . Let us define, for all natural numbers $m$ , $k$ with $1 \leq k \leq m$ $$\left[\begin{array}{ccc}m k\end{array}\right] =\frac{ 6^{(m)} 6^{(m-1)}\cdots 6^{(m-k+1)}}{6^{(1)} 6^{(2)}\cdots 6^{(k)}} .$$ Prove that for all $m, k$ , $\left[\begin{array}{ccc}m k\end{array}\right]$ is a natural number whose decimal representation consists of exactly $k(m + k - 1) - 1$ digits.