Prove equation with induction Prove by induction that for every natural number $n\in\mathbb{N}$ and every real number $x\in\mathbb{R}$:
$$(1+x)(1+x^2)(1+x^4)\cdot\,\dots\,\cdot(1+x^{2^{n-1}}) = 1 +x +x^2+\dots+(x^{2^n-1})$$
I proved for $n=1$ but i don't know how to continue with $(n+1)$
THANKS
 A: I think you made some mistake in the formula. We prove 
$$(1+x)(1+x^2)\cdots (1+x^{2^{n-1}}) = 1+x+x^2+x^3+\cdots + x^{2^n-1}.$$
For $n=1$ we get $x+1=x+1$ which holds.
Assume the statement holds for $n$. We prove it holds for $n+1$. So we have
$$(1+x)(1+x^2)\cdots(1+x^{2^{n-1}})(1+x^{2^{n}}) = (1+x+\cdots +x^{2^n-1})(1+x^{2^{n}}),$$
by our induction assumption. But
$$(1+x+\cdots +x^{2^n-1})(1+x^{2^{n}})=1+x+\cdots +x^{2^n-1} + x^{2^n}+x^{2^n+1} \cdots + x^{2^n+2^{n-1}}=1+x+\cdots+x^{2^{n+1}-1},$$
proving the statement.
A: I can't resist to show non purely inductive proof. Denote
$$
P=(1+x)(1+x^2)\cdot\ldots\cdot(1+x^{2^{n-1}})
$$
Note that
$$
\begin{align}
(1-x)P&=(1-x)(1+x)(1+x^2)(1+x^4)\cdot\ldots\cdot(1+x^{2^{n-1}})\\
&=(1-x^2)(1+x^2)(1+x^4)\cdot\ldots\cdot(1+x^{2^{n-1}})\\
&=(1-x^4)(1+x^4)\cdot\ldots\cdot(1+x^{2^{n-1}})\\
&=\ldots\\
&=(1-x^{2^{n-1}})(1+x^{2^{n-1}})\\
&=1-x^{2^n}
\end{align}
$$
Hence
$$
P=\frac{1-x^{2^n}}{1-x}=1+x+x^2+\ldots+x^{2^n-1}
$$
A: Here is a proof that doesn't seem to employ induction but uses the uniqueness of the binary representation of the non-negative integers (which may use induction).
Let $0\le m\lt2^n$ then there is only one way to get $x^m$ from the product
$$
(1+x)(1+x^2)(1+x^4)\dots(1+x^{2^{n-1}})
$$
that is by choosing $x^{2^k}$ in the factors where bit $k$ is $1$ in the binary representation of $m$ and choosing $1$ in the factors where bit $k$ is $0$. All coefficients in the factors are $1$ so the coefficient of $x^m$ in the product is $1$.
Therefore,
$$
(1+x)(1+x^2)(1+x^4)\dots(1+x^{2^{n-1}})=1+x+x^2+x^3+\dots+x^{2^n-1}
$$
