Base Case: m$_1$...m$_k$ is m$_1$. The hypothesis implies that p thus divides m$_1$. Thus, p divides at least one of the integers of the product m$_1$.
Recursive Step: Suppose that for at least one m$_k$, p divides m$_1$...m$_n$. By Bill Dubuque's hint, p divides m$_1$...m$_n$m$_j$ where j = (n + 1) implies that p divides m$_1$...m$_n$ or m$_j$. If it divides m$_1$...m$_n$, then by assumption for at least one m$_k$ it divides m$_1$...m$_n$. If it divides m$_j$, then for at least one m$_k$ it divides m$_j$. Thus, in either case for at least one m$_k$, p divides m$_1$...m$_n$m$_j$. Therefore, if for at least one m$_k$, p divides m$_1$...m$_n$, then for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.
Your set-up is pretty weird, because the 'inner' induction actually does not depend on the outer induction at all. That is, you don't do the 'inner' induction inside the 'outer' induction, but rather, you end up with two separate inductions:
The 'outer' induction shows that the claim holds for all $(n,n+1)$ (note: you'll have to use the base case $(1,2)$ if you do ... rather than say that you just take some 'arbitrary case')
The 'inner' induction takes some arbitrary $n$, and then shows, by induction, that the claim holds for any $j>n$.
Well ... notice that the latter is sufficient to prove the whole thing!
To be specific, with the 'inner' induction', you end up just do the following:
Take any arbitrary $n$
Now you show by the 'inner' induction that for all $j>n$, the claim holds
But since $n$ was arbitrary, the claim holds for all $n, j>n$
That is, you do a universal proof on $n$, and inside that you induct on $j$, and with that, the whole proof is done!
Indeed, the 'inner' induction is only 'inner' to the universal proof you do on $n$.
This means that the 'outer' induction in not necessary at all!
Indeed, you effectively end up showing all the $(n,n+1)$ cases through the base case of the inner induction. To be precise, when you show that for any arbitrary $n$, the base case of the inner induction holds, you have in effect shown that the claim holds for all $(n,n+1)$.
Best Answer
I think you're almost all the way there-- the main feedback I'd give is around your final statement:
$$9|10^{n+1}-1\implies9|(9k+1)\cdot10-1\implies9|90k+9$$
Above you're assuming what you want to prove. Try working backwards starting with what you know to get to what you want to prove. Here, try the following to get started:
$$10^n = 9k + 1 \implies 10^{n+1} = (9k + 1) \cdot 10 \implies 10^{n+1} - 1 = ... $$
N.B. Also be sure to declare what $k$ is when you use it.
I hope this helps!