Winding number of composition of maps If $f,g:S^{1}\rightarrow S^{1}$ maps, show that $N(f\circ g)=N(f)N(g)$, where $N(f)$ is the winding number of $f$.
We defined the winding number of $f$ to be $N(f)=\frac{1}{2\pi}(\tilde{f}(1)- \tilde{f}(0) )$ where $\tilde{f}$ is a lift of $f$. 
I tried to look for a lift of $f \circ g$ but couldn't manage to find anything that makes sense for this problem. 
Can anybody help me, please? 
Thank you in advance!

This may be a bit "cheating", but anyway... If $n = N(f)$ and $m = N(g)$, we know from the calculation of $\pi_1(S^1)$ that $f$ is homotopic to $z \mapsto z^n$ and $g$ is homotopic to $z \mapsto z^m$ (two maps with the same winding number are homotopic). But then $f \circ g$ is homotopic to $z \mapsto (z^m)^n = z^{mn}$, which has winding number $mn = N(f)N(g)$.
