<s>[INST] Find the number of zeroes in ${999\cdots99}^m$ 
Find the number of $\text{zeroes}$ in $\mathrm{999 \cdots 99}^{\mathrm{m}}$ for some given $m \in \mathbb{Z^+}$ provided that there are $\mathrm{n~many~9's}$.

I observed that when $m=3$, the zeroes increase by $1$ starting from $1$.
For $m=4$, starting from $2$, it increases by $2$ and for $m=5$, it increases by $2$ again but for $m=6$, it increases by $3$. 
I can only prove why ${(\underbrace{\mathrm{999 \cdots 99}}_{\text{n many}})}^3$ has $\text{n-1 zeroes}$. I am not stating the proof as it's done after expanding $(10^n-1)^3$. 
How can it he generalized? (The question itself is the generalized version)
 [/INST] Here are a few data points ($1 \le m,n \le 15$). It doesn't look very regular for low $n$.
\begin{array}{}
m \setminus n & 1 & 2 & 3 & 4 \\
1 & 0 \\
2 & n-1 \\
3 & n-1 \\
4 & 2(n-1) \\
5 & 2n-1 \\
6 & 0 & 3(n-2)+2 \\
7 & 0 & 3(n-2)+2 \\
8 & 1 & 4(n-2)+1 \\
9 & 1 & 1 & 4(n-2) \\
10 & 1 & 7 & 5(n-2) \\
11 & 2 & 1 & 5(n-3)+4 \\
12 & 0 & 6(n-2)+2 \\
13 & 0 & 1 & 2 & 6(n-3) \\
14 & 0 & 3 & 15 & 7(n-3)+15 \\
15 & 2 & 2 & 6 & 7(n-4)+9
\end{array}
</s>