Proof of the identity $F_{(n+1)k}=\frac{F_{2k}}{F_k}F_{nk}+(-1)^{k+1}F_{(n-1)k}\quad\forall k,n\in\mathbb{N}$ I'm reading the solution to an exercise I have been given in which they use the identity
$$F_{(n+1)k}=\frac{F_{2k}}{F_k}F_{nk}+(-1)^{k+1}F_{(n-1)k}\quad\forall k,n\in\mathbb{N}$$
where $F_n$ denotes the $n$th Fibonacci number. Is this a commonly known identity? Is there any way to obtain this identity in a relatively straightforward manner?

$$ \varphi = \frac{1+\sqrt5}{2}$$
$$ F_n = \frac{\varphi^n +(-1)^{n+1}\cdot \varphi^{-n}}{\sqrt5} $$
Example: $$ F_{13} = \frac{\varphi^{13} + \varphi^{-13}}{\sqrt5} = 233 $$
Then: $$ \frac{F_{2k}}{F_k}=\frac{\frac{\varphi^{2k} +(-1)^{2k+1}\cdot \varphi^{-2k}}{\sqrt5}}{\frac{\varphi^k +(-1)^{k+1}\cdot \varphi^{-k}}{\sqrt5}} = \frac{\varphi^{2k} +(-1)^{2k+1}\cdot \varphi^{-2k}}{\varphi^{k} +(-1)^{k+1}\cdot \varphi^{-k}} =$$
$$ \frac{\varphi^{2k} - (-\varphi)^{-2k}}{\varphi^{k} - (-\varphi)^{-k}} = \varphi^{k} + (-\varphi^{-1})^{k}$$
and set:
$$ \varphi^k = a $$
$$ (-\varphi^{-1})^k = b $$
$$ G_n = F_{kn} = \frac{a^n - b^n}{\sqrt5} $$
Test:
$$ G_{n+1} = \frac {G_2}{G_1}G_n-(-1)^kG_{n-1}$$
$$ G_{n+1} + (-1)^kG_{n-1} = \frac {G_2}{G_1}G_n$$
$$ \frac{a^{n+1} - b^{n+1}}{\sqrt5} + (-1)^k\frac{a^{n-1} - b^{n-1}}{\sqrt5} = (a+b)\frac{a^n - b^n}{\sqrt5}$$
$$ a^{n+1} - b^{n+1}+ (-1)^k(a^{n-1} - b^{n-1}) = (a+b)(a^n - b^n)$$
$$ a^{n+1} - b^{n+1}+ (-1)^k(a^{n-1} - b^{n-1}) = a^{n+1} - b^{n+1} + ba^n - ab^n$$
$$ (-1)^k(a^{n-1} - b^{n-1}) = ba^n - ab^n $$
$$ (-1)^k(a^{n-1} - b^{n-1}) = ab(a^{n-1} - b^{n-1}) $$
$$ (-1)^k = ab \lor a^{n-1} - b^{n-1} = 0 $$
But:
$$ ab = \varphi^k \cdot (-\varphi^{-1})^k = \varphi^k \cdot \varphi^{-k} \cdot (-1)^k =  (-1)^k $$
Demostrated.
