<s>[INST] 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.
 [/INST] Let the induction statement be such that 
$$P(n):\prod_{i=0}^{n} 1 + q^{2^i}=\dfrac{1-q^{2^{n+1}}}{1-q}$$
We show the base $(n=0)$ case, which states
$$P(0):\prod_{i=0}^{0} 1 + q^{2^0}=\dfrac{1-q^{2^{0+1}}}{1-q}$$
Working from the right hand side, we have 
$$\dfrac{1-q^{2^{0+1}}}{1-q}=\dfrac{1-q^{2}}{1-q}=\dfrac{(1-q)(1+q)}{1-q}=1+q=\prod_{i=0}^{0} 1 + q^{2^0}$$
With the base case verified, assume $P(n)$ is true and we must show $P(n+1)$ which states
$$P(n+1):\prod_{i=0}^{n+1} 1 + q^{2^i}=\dfrac{1-q^{2^{(n+1)+1}}}{1-q}$$
This should be straight forward starting from the left hand side.
$$\begin{align}
\prod_{i=0}^{n+1} 1 + q^{2^i}&=\Bigg(\prod_{i=0}^{n} 1 + q^{2^i}\Bigg)\cdot(1+q^{2^{n+1}})
\\
&=\Bigg(\dfrac{1-q^{2^{n+1}}}{1-q}\Bigg)\cdot(1+q^{2^{n+1}})
\\
&=\dfrac{1-q^{2^{n+1}}q^{2^{n+1}}}{1-q}
\\
&=\dfrac{1-q^{2^{n+1}+2^{n+1}}}{1-q}
\\
&=\dfrac{1-q^{2\cdot2^{n+1}}}{1-q}
\\
&=\dfrac{1-q^{2^{n+2}}}{1-q}
\\
&=\dfrac{1-q^{2^{(n+1)+1}}}{1-q}
\end{align}$$
Thus,
$$\prod_{i=0}^{n+1} 1 + q^{2^i}=\dfrac{1-q^{2^{(n+1)+1}}}{1-q}$$
And $P(n+1)$ is true for all $n \in \mathbb{N} \cup \{0\}$.
</s>