Prove that for any positive integer $n$ there exists an $n$-digit number that is divisible by $5^{n}$ and contains only the digits $5, 6, 7, 8, 9$.
My attempted solution:
Proof by induction.
The base case: $n=1$.
$m=5$ is divisible by $5^{1}$ and contains only the digits $5, 6, 7, 8, 9$.
The induction hypothesis: there exists an n-digit number $m=5^{n}k$ that is divisible by $5^{n}$ contains only the digits $5, 6, 7, 8, 9$.
The induction step: the $(n+1)$-digit number will be one of the following,
$5*10^{n} + m = 5^{n}(5*2^{n}+k)$
$6*10^{n} + m = 5^{n}(6*2^{n}+k)$
$7*10^{n} + m = 5^{n}(7*2^{n}+k)$
$8*10^{n} + m = 5^{n}(8*2^{n}+k)$
$9*10^{n} + m = 5^{n}(9*2^{n}+k)$
Then, I'm not sure how to proceed from here. I know I have to apply my number theory knowledge here. But not sure how.