<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] We will start off with the basis step using the base value of $n = 0$.
Basis step
$\prod_{i=0}^{0} (1 + q^{2^i}) = \dfrac{1 - q^{2^{n + 1}}}{1 - q}$
=> $1 + q^{2^0} = \dfrac{1 - q^{2^{0 + 1}}}{1 - q}$
=> $1 + q = \dfrac{1 - q^2}{1 - q}$
=> $(1 + q)(1 - q) = 1 - q^2$
=> $1 - q^2 = 1 - q^2$
The basis step is hereby proved by equality and we move forward to the inductive step.
Inductive step
If it is the case that $\prod_{i=0}^{n} (1 + q^{2^i}) = \dfrac{1 - q^{2^{n + 1}}}{1 - q}$ is true, then is $\prod_{i=0}^{k} (1 + q^{2^i}) = \dfrac{1 - q^{2^{k + 1}}}{1 - q}$ also true for any $0 \leq k$, which forms our hypothesis.
For this equality to be true, our hypothesis must also be true for $k + 1$.
$\prod_{i=0}^{k} (1 + q^{2^i})(1 + q^{2^{k + 1}}) = \dfrac{1 - q^{2^{k + 2}}}{1 - q}$
We use our hypothesis for a substitution:
$(\dfrac{1 - q^{2^{k + 1}}}{1 - q})(1 + q^{2^{k + 1}}) = \dfrac{(1 - q^{2^{k + 1}})(1 + q^{2^{k + 1}})}{1 - q} = \dfrac{1 - (q^{2^{k + 1}})^2}{1 - q} = \dfrac{1 - q^{2^{k + 1} * 2}}{1 - q} = \dfrac{1 - q^{2^{k + 2}}}{1 - q}$, which completes our proof.
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