In this paper of A. T. Benjamin and S. S. Plot https://www.fq.math.ca/Papers1/46_47-1/Benjamin_11-08.pdf there's the proof that the following coefficient is an integer:
$$\binom{n}{k}_F = \frac{F_nF_{n-1}\dots F_{n-k+1}}{F_kF_{k-1}\dots F_1}$$
where $F_n = F_{n-1} + F_{n-2}$, for $n \ge 2$, and $F_0=0$, $F_1=1$. It is clear that if $f_n$ is the $n$-th Fibonacci's number, then $f_n = F_{n+1}$.
A well known combinatorial interpretation of these numbers is that $F_n = f_{n-1}$ counts the number of tilings of a $(n-1) \times 1$ board, with squares $1 \times 1$ and dominoes $2 \times 1$. Hence $F_nF_{n-1}\dots F_{n-k+1}$ enumerates the number of simultaneous tilings, in squares and dominoes, of $(n-1) \times 1$, $(n-2)\times 1$, $\dots$, $(n-k)\times 1$ boards.
To show that $\binom{n}{k}_F$ is an integer, they "cut away" a disjointed covering in squares and dominoes of $(k-1) \times 1$, $(k-2)\times 1$, $\dots$, $1 \times 1$ boards (counted by $F_kF_{k-1}\dots F_1$) from those described above representing the numerator.
The procedure is not that hard: first of all we are looking for a $(k-1)\times 1$ covering, starting by the $(n-1)\times 1$ board. If a domino doesn't cover the cells $k-1$ and $k$, we can break the board at $k-1$ finding what we wanted. It can happen in $f_{k-1}f_{n-k} = F_kF_{n-k+1}$ ways, by considering the number of coverings of the two independent boards we got breaking the $(n-1)\times 1$ at $k-1$.
Naming $T_i$ a tailing of $(n-i)\times 1$. If a domino is laid on the $k-1$ and $k$ cells, that can happen in $f_{k-2}f_{n-k-1} = F_{k-1}F_{n-k}$ ways, by throwing away that domino and taking into account the new two boards $(k-2)\times 1$ and $(n-k-1)\times 1$ disjointed, then we eliminate the first $k$ cells obtaining a $(n-k-1) \times 1$ tiling, that is $T_{k+1}$. We follow this procedure on the next $(n-2)\times 1$, $(n-3)\times 1$, $\dots$, since we find a tiling $T_{x_1}$that is breakable at $k-1$, which exists as $T_{n-k+1}$ has length $k-1$. So $1 \le x_1 < n$.
Iterating this method (looking for $(k-2)\times 1$ by starting from $T_{x_1+1}$, etc.) we find a disjointed covering $T_1, \dots, T_k$ of $(k-1)\times 1, \dots, 1\times1$. Then they write:
"Following this procedure, we have, for $1 \le x_1 < x_2 < ··· < x_{k-1} \le n$, the number of tilings $T_1, T_2, \dots , T_k$ that lead to finding a tiling of length $k − i$ at the beginning of tiling $T_{x_i}$ is
$f_{k-2}^{x_1-1}f_{k-1}f_{n-x_1 - (k-1)}f_{k-3}^{x_2-x_1-1}f_{k-2}f_{n-x_2 - (k-2)}\dots f_{0}^{x_{k-1}-x_{k-2}-1}f_{1}f_{n-x_{k-1} - 1}$"
I haven't fully understood this result yet. The process to obtain $(k-1)\times1$ sees the contribute of $f_{k-2}$ any time we can't break the board at $k-1$, which happens for $x_1-1$ times, so we have $f_{k-2}^{x_1-1}$. I assume that $f_{k-1}$ derives from the case in which we can break it and stop the iteration for $(k-1)\times 1$ (?), then $f_{n-x_1 - (k-1)}$ how does it come up? Any explanation would be appreciated, thank you.
EDIT
Maybe I figured it out and it was easier that I thought: when the process stops at $x_1$ we are considering a $n-x_1$ row, so $f_{k-1}f_{n-x_1 - (k-1)}$ is the number of independent tilings of the that board broken at $k-1$!
The proof of this is elementry, and considerably simpler than shown.
$F_n$ is a repunit in the base $F_n =a^n-a^{-n}$ where $a=\sqrt{-1/4}+\sqrt{-5/4}$.
This means that there is a factor for each integer $m$, such that $f_m\mid F_n$ iff $m\mid n$. Also, $F_n = \prod f_m \text{ over } m\mid n$.
So all it is to be shown is that there are at least the same number of multiples of $m$ from $1$ to $k$ in the numerator. This will happen automatically if the numerator is a product of consecutive values.