Fermat Numbers Theorem Proof by Induction

Question

Let $F(n)$ be the $n$th Fermat number.

I wish to prove that:

$F(n+1) - 2 = F(0) * F(1) * F(2) * \cdots * F(n)$

For this I used proof by induction and my steps were as follows:

For n=1: LHS = $F(2) -2 - 15$ and RHS = $F(0) * F(1) = 15$

LHS = RHS => true for $n=1$

Assume true for $n=k$:

$F(k+1) - 2 = F(0) * F(1) * ... * F(k)$

Now, consider RHS for $n=k+1$:

$F(0) * F(1) * ... * F(k) * F(k+1) = {F(k+1) - 2} * F(k+1) = (2^{2^{k+1}} - 1) * (2^{2^{k+1}} + 1)$

This is the point I am stuck with as I am not sure what I get when I square $2^{2^{k+1}}.$

I would appreciate any help! Thank you!

Answer 1

As you surly know, you need to use $(a-b)\times(a+b) = a^2 - b^2$ with $a = 2^{2^{k+1}}$ and $b=1$. Now we have \begin{equation} \left(2^{2^{k+1}}\right)^2 = 2^{2^{k+1}\times 2} = 2^{2^{k+2}} \end{equation} and $a^2 - b^2 = 2^{2^{k+2}}-1 = F(k+2) - 2$.

Answer 2

Using the laws of exponents $$2^{(2^{k+1})}\cdot 2^{(2^{k+1})}=2^{(2^{k+1}+2^{k+1})}=2^{(2^{k+2})}$$ which is just what you are looking for