n-step composition of a homographic function ; proof by induction Prove that $f_n (l) = \frac{l}{l(n+1)+1}$ for all positive integers n.
If $f_{n+1}=f_0 \circ f_n$ and $f_{0} (l)=\frac{l}{l+1}$ for $n= 0,1,2,3....$ 
This proof has to be solved with the method of induction. Now, I ran into a problem with base cases. For the $f_n (l)$ the base case is 1 or n=1, but the base case for $f_{n+1}$ is $n=0$. So I don't know how to take this further or I might be completely thinking wrong. 
 A: The base case is in fact $n=0$ for which you need to prove that $$f_1=f_0Of_0$$since $$f_0Of_0(l)={f_0(l)\over f_0(l)+1}={{l\over l+1}\over {l\over l+1}+1}={l\over l+l+1}={l\over 2l+1}=f_1(l)$$hence the induction is complete.
A: The coefficients you are looking for are those you find in the matrix power :
$$\begin{pmatrix}1&0\\1&1\end{pmatrix}^n=\begin{pmatrix}1&0\\n&1\end{pmatrix}\tag{1}$$
Therefore, your final composition of functions is 
$$f(\ell):=\dfrac{\ell}{n\ell+1}\tag{2}$$
This is due to the classical correspondence :
$$f(x):=\dfrac{ax+b}{cx+d} \ \ \leftrightarrow \ \ M_f=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
which induces a homomorphism between composition of "homographic" functions and matrix product (up to a factor)
$$f_1 \circ f_2  \ \ \leftrightarrow \ \ M_{f_1}M_{f_2}$$
Proof : (I take $x$ instead of $\ell$, not usual).
$$\dfrac{a \tfrac{ex+f}{gx+h} +b}{c\tfrac{ex+f}{gx+h} +d}=\dfrac{a(ex+f)+b(gx+h)}{c(ex+f)+d(gx+h)}=\dfrac{(ae+bg)x+(af+bh)}{(ce+dg)x+(cf+dh)}$$ 
where you recognize the coefficients of the product of matrices :
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}e&f\\g&h\end{pmatrix}.$$ 
Caution : The last equality in (1) necessitates a (rather easy) reasoning by induction.
