Prove by induction ${\prod\limits_{k=1}^{n}(1+x^{2k-1}}) = \frac{1-x^{2n}}{1-x} .$
We start verifying for $n = 1$:
${\prod\limits_{k=1}^{1}(1+x^{2k-1}})= $$\frac{1-x^{2(1)}}{1-x}$
${\prod\limits_{k=1}^{1}(1+x^{2k-1}})= $ $\frac{1-x^{2}}{1-x}$
${\prod\limits_{k=1}^{1}(1+x})= $ $\frac{{(1+x)(1-x)}}{1-x}$
${\prod\limits_{k=1}^{1}(1+x})= $ $(1+x)$
Then we accept that the equation is valid for n and we have to n + 1.
${\prod\limits_{k=1}^{n+1}(1+x^{2k-1}})= $$\frac{1-x^{2(n+1)}}{1-x}$
${\prod\limits_{k=1}^{n+1}(1+x^{2k-1}})= $$\frac{1-x^{(2n+2)}}{1-x}$
Now the problem is that I have come to develop up to the part where the multiplication of
$(\frac{1-x^{2n}}{1-x})(1+x^{2(n+1)-1})$
I did the math but it does not give me the expected result. Maybe I put the operation wrong.
From that point I have no idea how to continue.
Proof by induction: clearly true when $n=1$.
The inductive hypothesis is \begin{eqnarray*} \prod_{k=1}^{n}(1+x^{2^{k-1}}) =\frac{1-x^{2^n}}{1-x}. \end{eqnarray*} The inductive step is \begin{eqnarray*} \prod_{k=1}^{n+1}(1+x^{2^{k-1}}) =(1+x^{2^{n}}) \prod_{k=1}^{n}(1+x^{2^{k-1}})= (1+x^{2^{n}})\frac{1-x^{2^n}}{1-x}=\frac{1-x^{2^{n+1}}}{1-x}. \end{eqnarray*}