Given $x\odot y=x+y+xy$, prove $x\odot x\odot\ldots\odot x= (1+x)^n -1$ 
Given $x\odot y=x+y+xy$, prove that 
  $$\underbrace{x\odot x\odot\ldots\odot x}_{n\text{ times}}= (1+x)^n -1$$ for all $n\in \Bbb{N}$ and $x \in\Bbb R\setminus\{-1\}$.

I have tried to use the binomial theorem on $(1+x)^n$ but was unable to simplify the nested function. Also tried mathematical induction but that didn't work out. The nested function is still stumping me as I haven't figured out how to generalize it for $n$ terms. Also some information on how to deal with nested functions would be greatly appreciated. 

By induction:
$$x_1 = x$$
$$x_n = x \odot x_{n-1} = x+x_{n-1} + xx_{n-1} \Rightarrow \\
x_n = (1+x)x_{n-1} + x.$$
Then:
$$x_n = (1+x)\left[(1+x)x_{n-2} + x\right] + x = \\
= (1+x)^2x_{n-2} + (1+x)x + x = \\
= (1+x)^3x_{n-3} + (1+x)^2x + (1+x)x + x \Rightarrow\\
x_n = (1+x)^{h}x_{n-h}+x\sum_{k=0}^{h-1} (1+x)^k.$$
For $h=n-1$:
$$x_n = (1+x)^{n-1}x_{1}+x\sum_{k=0}^{n-2} (1+x)^k \Rightarrow \\
x_n = (1+x)^{n-1}x + x\frac{1-(1+x)^{n-1}}{1-(1+x)} \Rightarrow \\
x_n = (1+x)^{n-1}x - (1-(1+x)^{n-1}) \Rightarrow \\
x_n = (1+x)^{n-1}(1+x) - 1 \Rightarrow\\
x_n = (1+x)^n - 1.$$
