"(5) was intended to assume that $n \in \mathbb ..."

(5) was intended to assume that $n \in \mathbb R^{\ge 1},$ or possibly $\in \mathbb R^{\ge 0}$ if you want an easy way to prove (6). In that case, how does (8) not follow from (5)? (If $f(x^y)=yf(x)$ in general, then $f(b^n)=nf(b)$ and $f(b)=1 \implies f(b^n)=n,$ unless I'm missing something.)