Fermat number proof

Question

I want to show that $F_n≡ 2 (mod F_i)$ for $i < n$, where $F_n$ is a Fermat number. My thoughts so far:

$2^{2^n}+1≡0 (mod F_n)$, so it follows that $2^{2^n}≡-1(modF_n)$. Squaring and then adding 1 to both sides gives $2^{2^{n+1}}+1=F_{n+1}≡2(mod F_n)$. I think this shows that $F_n≡2(modF_i)$ where $i=n-1$, but I'm not quite sure how to show that $F_n ≡2(mod F_i)$ for any $i<n$. I would appreciate some help!

Also, by induction we have $F_0F_1...F_{n-1}=F_n-2$:

Supposing the above is true, we have $F_0F_1...F_n=(F_n-2)F_n=(2^{2^n})^2+2(2^{2^n})+1-2F_n=2^{2^n+1}-1=F_{n+1}-2.$

Answer 1

Just keep on squaring to get that for every $k>0$, $$2^{2^{n+k}}\equiv1\bmod F_n$$ The result then follows as before by adding $1$.

Answer 2

It boils down to: Fermat's minus $1$ are powers of $2$ so obey $\color{#0a0}{\text{law of exponents}}$ $$\begin{align} &\overbrace{F_{n+k}-1^{\phantom{|^{|}}}\!\!\!}^{\textstyle\ \ \ \ \ \color{#0a0}{2^{\large 2^{n+k}}}}\, =\, \overbrace{(\color{#c00}{F_{n}}-1)^{\large 2^k}}^{\textstyle\ \ \ \ \ \color{#0a0}{(2^{\large 2^n})^{\large 2^k}}}\\[.2em] \Rightarrow\, \bmod \color{#c00}{F_n}\!:\,\ &F_{n+k}-1\,\equiv\ \ (\color{#C00}0\,-\,1)^{\large 2^k}\!\!\underset{k>0^{\phantom |}\!}\equiv\! 1 \end{align}$$