Proving that $\log_b(r^s) = s\log_b(r)$ The Question
If $b,r,s \in \mathbb{R^+}$ prove that $\log_b(r^s) = s\log_b(r)$
My Work
1) $\log_b(r^s)$ 
2) $s$ can be expressed as the sum of an integer part $n$ and a real part $m$: $s = m + n$
3) $\log_b(r^{n+m})$
4) $\log_b(r^nr^m)$
5) $\log_b(r^n) + \log_b(r^m)$
6) $\log_b(r\cdot r\cdot r \cdots r) + \log_b(r^m)$
7) $\log_b(r) + \log_b(r) + \log_b(r) + \cdots + \log_b(r) + \log_b(r^m)$
8) $n\log_b(r) + \log_b(r^m)$
Where I am Having Trouble
I'm having trouble getting the $m$ in front of the second summand which I feel is necessary for the theorem. How can I finish off this proof? I was given this rule as a secondary school student with no proof and would like to have it explained. 

Take $\log_b (r^s):=F(s)$ ($r$ an arbitrary constant) and note that $F(0)=\log_b (1)=0$
Show that:
$F(s)=F(s-1)+\log_b r$
Hence $\frac{F(s)-F(s-1)}{s-(s-1)}=\log_b (r)$ and $F$ is of slope $\log_b (r)$:
$$F(s)=(\log_b r)s+c$$
But $F(0)=0$ gives $c=0$.
If we proved it for all positive arbitrary constants $r>0$, then the equation must hold for all $r>0$.
