For $ m\in\mathbb{N}$, be $ f(m)=m\cdot 10^{r}$ with $ r$ the unique integer for which $ 1\le f(m)<10$ (i.e. the first digit of $ m$). Then for $ 2\le i\le9$ we have $ i<f((N+i)!)<i+1$, such that \[ 1 < f(N+i) = f\left(\frac{(N+i)!}{(N+i-1)!}\right) <\frac{i+1}{i-1}\le 3.\] So $ f(N+i)<f(N+i-1)$ and thus $ 1 < f(N+2) < f(N+3) < ... < f(N+9) <\frac{5}4$. Since obviously $ f(ab)\le f(a)f(b)$, is \[ f((N+4)!)\le f((N+1)!)f(N+2)f(N+3)f(N+4)< 2\left(\frac{5}4\right)^{3}< 4,\] contradiction.

Peter wrote: ...... Since obviously $ f(ab)\le f(a)f(b)$........ Excuse me. I don't undertand this one. $ f(7 \times 13) > f(7)f(13)$

No, $ f(7.13)=9\cdot8=7\cdot1.3=f(7)\cdot f(13)$? I don't understand the objection. Either we have LHS=RHS or we have LHS=RHS/10, so we have the stated inequality?