number theory - $7^n$ contains a block of consecutive zeroes

Prove that there is a positive integer $n$ such that the decimal representation of $7^n$ contains a block of at least $m$ consecutive zeros, where $m$ is any given positive integer.

I will prove it more generally for any prime $p$. It is sufficient to find an $n$ such that $p^n$ begins with the number $100\ldots 0$ which has exactly $m$ zeroes. Thus, we are looking for $n$ and $k$ with $k < m$ such that $10^m 10^k \leq p^n < 10^k(10^m+1)$. This is equivalent to $m \leq n\log_{10}{p}-k < \log_{10}(10^m+1)$.

Where do I go from here?