How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$? 
Prove the identity
  \begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
  for each nonnegative integer $n$.

To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$
If $n=1$ then I have next expressions
$(1+q)(1+q^2) = \frac{1-q^{2^{1+1}}}{1-q} = $
$ = (1+q)(1+q^2) = \frac{1-q^{4}}{1-q}$ 
Well played and calculated =(.Speak me where i s mistaked.

For the base case $n=1$, prove that $$(1+q)(1+q^2)=\frac{1-q^4}{1-q}\iff \underbrace{(1-q)(1+q)}_{=(1-q^2)}(1+q^2)=1-q^4$$
Which is true since $(a-b)(a+b)=a^2-b^2$.
For the inductive step $\color{blue}{\text{assume that the equation holds for some $n$}}$. Therefore
$$\begin{align*}(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n+1}})&=\color{blue}{(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n}})}\cdot (1+q^{2^{n+1}})\\ &=\color{blue}{\frac{1-q^{2^{n+1}}}{1-q}}\cdot (1+q^{2^{n+1}})\\ &=\frac{1-q^{2^{n+2}}}{1-q}\end{align*}$$
Alternatively
Although induction is usually a nice approach when it comes to proving an equation $\forall n\in\Bbb N$, there is even a nicer way to prove your equation. Observe simply that $$\begin{align*}&(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n}})=\frac{1-q^{2^{n+1}}}{1-q}\\\iff& (1-q)(1+q)(1+q^2)...(1+q^{2^{n}})=1-q^{2^{n+1}}\end{align*}$$ which can be proven using repetidly the fact that $(a-b)(a+b)=a^2-b^2$ Therefore $$\begin{align*}(1-q)(1+q)(1+q^2)...(1+q^{2^{n}})&=(1-q^2)(1+q^2)...(1+q^{2^{n}})\\&=(1-q^4)(1+q^4)...(1+q^{2^{n}})\\&=\;...\\&=(1-q^{2^{n}})(1+q^{2^{n}})\\&=(1-q^{2^{n+1}})\end{align*}$$
Which is what we wanted to prove.
