$ f(n)$ is last digit different from 0 of $ n!$ Suppose this sequence period T Thwn exist $ m,k$ such that : $ 10^m\equiv 10^k(\mod T)$ We has $ f(10^k) = f(10^{k - 1})$ $ f(10^k + 10^k - 10^m) = f(10^k + 10^k - 10^m - 1)$ So $ f(2.10^k - 10^m) = f(2.10^k - 10^m - 1)$ Other: $ (2.10^k - 10^m)! = (2.10^k - 10^m - 1)!(2.10^k - 10^m)$ The last digit different from 0 of $ 2.10^k - 10^m$ is 9 So $ f(2.10^k - 10^m) = 9f(2.10^k - 10^m)$ So $ f(2.10^k - 10^m - 1)\equiv 9f(2.10^k - 10^m - 1)(\mod 10)$ so $ f(2.10^k - 10^m - 1) = 5$ But $ v_{2}(2.10^k - 10^m - 1) > v_{5}(2.10^k - 10^m - 1)$ So $ f(2.10^k - 10^m - 1)$ can not be 5 (tradition). So the sequence can not period.