- \begin{align}
F_{nm} = \sum_{k=0}^{m} \binom{m}{k} F_{n-1}^{k} F_{n}^{m-k} F_{m-k}.
\end{align}
This paper proves that
$$F_{kn+c}=\sum_{i=0}^{k}\binom kiF_{c-i}F_n^iF_{n+1}^{k-i}$$
You can set $c=k$ with $n\to n-1$.
- \begin{align}
F_{2^{n}} = \prod_{i=0}^{n-2} a_{i}, \hspace{3mm} a_{i}= a_{i-1}^{2} -2
\end{align}
From $a_{i}= a_{i-1}^{2} -2$ with $a_0=F_{2^2}=3$, one can prove that $a_i=L_{2^{i+1}}$ where $L_j$ is Lucas numbers. (You can use the fact that $L_{2n}=L_n^2-2(-1)^n$.)
This paper proves in Lemma 2 that
$$F_{2^{n}} = \prod_{i=0}^{n-2} L_{2^{i+1}}$$
- \begin{align}
F_{n+1} = 2^{n} \sqrt{\prod_{k=1}^{n} \cos^{2} \frac{k \pi}{n+1} + \frac{1}{4} }
\end{align}
This answer proves that
$$F_n^2 = \prod_{k=1}^{n-1}\left(3 + 2\cos\left(\frac{2\pi k}{n}\right)\right)$$
You can use $\cos 2A=2\cos^2A-1$ with $n\to n+1$.
- \begin{align}F_{n} = F_{n+2}-1 + F_{n+1}^{4} + 2(F_{n+1}^{3} F_{n+2}) - (F_{n+1}F_{n+2})^{2} - 2F_{n+1}F_{n-2}^{3} + F_{n+2}^{4}. \end{align}
I think you have typos. It should be
$$F_{n} = F_{n+2}-1 + F_{n+1}^{4} + 2(F_{n+1}^{3} F_{n+2}) - (F_{n+1}F_{n+2})^{2} - 2F_{n+1}F_{n\color{red}+2}^{3} + F_{n+2}^{4}\color{red}{-F_{n+1}}$$
Let $F_n=a,F_{n+1}=b$. Then, since $F_{n+2}=a+b$ and $\color{red}{b^2}=ab+a^2+(-1)^n$ (which is equivalent to $F_{n+1}F_{n-1}=F_n^2+(-1)^n$, and see here), we have
$$\begin{align}RHS&=a^4+a-1+2a^3b
\\&\qquad +\color{red}{b^2}( - a^2- 2 a b + \color{red}{b^2})
\\&=a^4+a-1+2a^3b
\\&\qquad +(ab+a^2+(-1)^n)(-ab+(-1)^n)
\\&=a+a^2(\underbrace{a^2+ a b + (-1)^n-\color{red}{b^2}}_{=0})
\\&=a\end{align}$$
- \begin{align} \frac{F_{n+3}^{2} - F_{n+1}^{2}}{F_{n+2}} = F_{n+3} + F_{n+1}. \end{align}
We have
$$LHS=\frac{(F_{n+3}+F_{n+1})(F_{n+3}-F_{n+1})}{F_{n+2}}=RHS$$
since $F_{n+3}-F_{n+1}=F_{n+2}$.
- \begin{align} F_{2n} = F_{n+2}^{2} - F_{n+1}^{2} - 2F_{n}^{2}. \end{align}
We have
$$\begin{align}RHS&=(F_{n+1}+F_n)^2-F_{n+1}^2-2F_n^2
\\&=2F_nF_{n+1}-F_n^2
\\&=F_n(F_{n+1}+F_{n+1}-F_n)
\\&=F_n(F_{n+1}+F_{n-1})\end{align}$$
Now, see here.
- \begin{align} F_{n} = \frac{F_{2n+k+1} - F_{n+1} F_{n+k+1}}{F_{n+k}}, \hspace{3mm} k\geq 0. \end{align}
It is sufficient to prove
$$F_{n+m} = F_nF_{m+1} + F_{n-1}F_m$$
(see for example here)
- \begin{align} F_{n+10} = 11F_{n+5} + F_{n}\end{align}
Using $F_N=F_{N-1}+F_{N-2}$, we have
$$\begin{align}F_{n+10}&=F_{n+9}+F_{n+8}
\\&=(F_{n+8}+F_{n+7})+(F_{n+7}+F_{n+6})
\\&=F_{n+8}+2F_{n+7}+F_{n+6}
\\&=(F_{n+7}+F_{n+6})+2(F_{n+6}+F_{n+5})
\\&\qquad +(F_{n+5}+F_{n+4})
\\&=F_{n+7}+3F_{n+6}+3F_{n+5}+F_{n+4}
\\&=(F_{n+6}+F_{n+5})+3(F_{n+5}+F_{n+4})
\\&\qquad +3F_{n+5}+(F_{n+3}+F_{n+2})
\\&=F_{n+6}+7F_{n+5}+3F_{n+4}+F_{n+3}+F_{n+2}
\\&=(F_{n+5}+F_{n+4})+7F_{n+5}+3F_{n+4}
\\&\qquad +(F_{n+5}-F_{n+4})+(F_{n+1}+F_{n})
\\&=9F_{n+5}+3F_{n+4}+F_{n+1}+F_n
\\&=9F_{n+5}+3F_{n+4}+F_{n+3}-F_{n+2}+F_n
\\&=9F_{n+5}+3F_{n+4}+(F_{n+5}-F_{n+4})
\\&\qquad -(F_{n+4}-F_{n+3})+F_n
\\&=10F_{n+5}+\underbrace{F_{n+4}+F_{n+3}}_{=F_{n+5}}+F_n
\\&=11F_{n+5}+F_n\end{align}$$
- \begin{align} F_{n} = 4F_{n-3} + F_{n-6}, \hspace{3mm} n\geq 6\end{align}
Using $F_N=F_{N-1}+F_{N-2}$, we have
$$\begin{align}F_n&=F_{n-1}+F_{n-2}
\\&=(F_{n-2}+F_{n-3})+(F_{n-3}+F_{n-4})
\\&=F_{n-2}+2F_{n-3}+F_{n-4}
\\&=(F_{n-3}+F_{n-4})+2F_{n-3}+(F_{n-5}+F_{n-6})
\\&=3F_{n-3}+\underbrace{F_{n-4}+F_{n-5}}_{=F_{n-3}}+F_{n-6}
\\&=4F_{n-3}+F_{n-6}\end{align}$$
