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)
 A: There isn't a simple answer as $m$ gets larger.  The proof for $m=3$ is instructive.  You write $(10^n-1)^3=(10^{3n}-3\cdot 10^{2n})+(3\cdot 10^n-1)$ and note that the places from $10^{n+1}$ through $10^{2n-1}$ will be zero, which is $n-1$ of them.  This works well for $m=4$, were we get $(10^n-1)^4=(10^{4n}-4\cdot 10^{3n})+(6\cdot 10^{2n}-4\cdot 10^n)+1$ and the places from $10^1$ to $10^{n-1}$ and from $10^{2n+1}$ to $10^{3n-1}$ are all zero for a total of $2n-2$.  The problem comes that the binomial coefficients start to carry.  At $m=5$ you have a couple coefficients that are $10$, but that just shifts things over.  At $m=6$ the middle one is $20$ and the ones next to it are $15$ so you don't get nice blocks of digits any more, though you do get blocks with no zeros.  Once $m=7$ you get a stray zero in the first block of non-zero digits and the problem only gets worse.  You can still chase the size of the blocks of zeros for a while, but it will get harder as you go up.  
You could make the general rule that for large enough $n$ the number of zeros is $k(m)+n\lfloor \frac m2 \rfloor$, but would have evaluate $k(m)$ by hand as $m$ gets larger and you need $n$ large enough that the blocks of zeros do not interfere with each other.
A: 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}
