p is without a common factor of 100000.Prove that power of p decimal representation ends with numbers group 00001.
Also prove that every natural nubmer n exist natural number k, that k power of p decimal representation ends with n zeros and number 1.
By the pigeon-hole principle, we find $k_1<k_2$ with $p^{k_1}\equiv p^{k_2}\pmod{10^{n+1}}$. As $p$ is coprime to $10$, $pa\equiv pb\pmod{10^{n+1}}$ implies $a\equiv b\pmod{10^{n+1}}$. So here we can ultimately conclude $p^0\equiv p^{k_2-k_1}\pmod{10^{n+1}}$, as desired.