Hello Ruslan,
First, we want to show the base case $P_1$ is true:
$$\prod_{k=0}^1\left(1+x^{2^k} \right)=\frac{1-x^{2^{1+1}}}{1-x}$$
$$\left(1+x^{2^0} \right)\left(1+x^{2^1} \right)=\frac{1-x^{2^2}}{1-x}$$
$$\left(1+x \right)\left(1+x^2 \right)=\frac{1-x^4}{1-x}=\frac{\left(1+x^2 \right)\left(1-x^2 \right)}{1-x}=\frac{\left(1+x^2 \right)(1+x)(1-x)}{1-x}=\left(1+x^2 \right)(1+x)$$
Thus, the base case is true. Hence, we state the given hypothesis:
$$\prod_{k=0}^n\left(1+x^{2^k} \right)=\frac{1-x^{2^{n+1}}}{1-x}$$
As our induction step, we may multiply both sides by $$\left(1+x^{2^{n+1}} \right)$$ to obtain:
$$\prod_{k=0}^n\left(1+x^{2^k} \right)\cdot\left(1+x^{2^{n+1}} \right)=\frac{1-x^{2^{n+1}}}{1-x}\cdot\left(1+x^{2^{n+1}} \right)$$
On the left side, incorporate the new factor into the product, and on the right carry out the indication multiplication:
$$\prod_{k=0}^{n+1}\left(1+x^{2^k} \right)=\frac{1-\left(x^{2^{n+1}} \right)^2}{1-x}$$
On the right apply the property of exponents $$\left(a^b \right)^c=a^{bc}$$ to obtain:
$$\prod_{k=0}^{n+1}\left(1+x^{2^k} \right)=\frac{1-x^{2\cdot2^{n+1}}}{1-x}$$
Now, on the right apply the property of exponents $$a\cdot a^b=a^{b+1}$$ to obtain:
$$\prod_{k=0}^{n+1}\left(1+x^{2^k} \right)=\frac{1-x^{2^{(n+1)+1}}}{1-x}$$
We have derived $P_{n+1}$ from $P_{n}$ thereby completing the proof by induction.