Given $a_1 =1,a_2=1, a_{n+2}=a_{n+1}+a_n$, prove that for $n\geq 2$, all coefficients of polynomial $$A(x)=\prod_{k=2}^{n+1}(1-x^{a_{k}})$$ are $0$, $1$, or $-1$.
I tried induction. I don't think it works. The hypothesis will be too weak. I think if we want to prove it by induction, we need to prove a stronger proposition, but I can't find it.
Yufei Zhao, The coefficients of a truncated Fibonacci power series, Fibonacci Quarterly, $\mathbf{46/47}\,(2008/2009)$, no. $1$, $53$-$55$, has the following clever inductive proof, which does indeed use a stronger hypothesis.
Say that a polynomial is timid if each of its coefficients is $-1,0$, or $1$; we want to show that $A_n(x)$ is timid for each $n\ge 1$, where
$$A_n(x)=\prod_{k=2}^{n+1}\left(1-x^{F_k}\right)\,.$$
We will use auxiliary polynomials $B_n(x)$ and $C_n(x)$ for $n\ge 1$:
$$\begin{align*} B_1(x)&=1-x^{F_2}-x^{F_3}=1-x-x^2\\ C_1(x)&=1+x^{F_1}-x^{F_3}=1+x-x^2\\ B_n(x)&=A_{n-1}(x)\left(1-x^{F_{n+1}}-x^{F_{n+2}}\right)\text{ for }n\ge 2\\ C_n(x)&=A_{n-1}(x)\left(1+x^{F_n}-x^{F_{n+2}}\right)\text{ for }n\ge 2\,. \end{align*}$$
Clearly $A_1(x)=1-x$, $B_1(x)$, $C_1(x)$, $A_2(x)=1-x-x^2+x^3$, $B_2(x)=1-x-x^2+x^4$, and $C_2(x)=1-x^2-x^3+x^4$ are timid. Suppose that $n\ge 3$, and $A_k(x),B_k(x)$, and $C_k(x)$ are timid for $1\le k<n$; we want to show that $A_n(x),B_n(x)$, and $C_n(x)$ are timid.
$$\begin{align*} A_n(x)&=A_{n-3}(x)\left(1-x^{F_{n-1}}\right)\left(1-x^{F_n}\right)\left(1-x^{F_{n+1}}\right)\\ &=A_{n-3}(x)\left(1-x^{F_{n-1}}-x^{F_n}\color{red}{-x^{F_{n+1}}+x^{F_{n-1}+F_n}}+x^{F_{n-1}+F_{n+1}}+x^{\color{blue}{F_n}+F_{n+1}}-x^{F_{n-1}+F_n+F_{n+1}}\right)\\ &=A_{n-3}(x)\left(1-x^{F_{n-1}}-x^{F_n}+x^{F_{n-1}+F_{n+1}}+x^{\color{blue}{F_{n-2}+F_{n-1}}+F_{n+1}}-x^{F_{n-1}+F_n+F_{n+1}}\right)\\ &=A_{n-3}(x)\left(1-x^{F_{n-1}}-x^{F_n}\right)+x^{F_{n-1}+F_{n+1}}A_{n-3}(x)\left(1+x^{F_{n-2}}-x^{F_n}\right)\\ &=B_{n-2}(x)+x^{F_{n-1}+F_{n+1}}C_{n-2}(x)\,. \end{align*}$$
Now
$$\deg B_{n-2}(x)=\sum\limits_{k=2}^{n-2}F_k+F_n=(F_n-F_1-1)+F_n=2F_n-2\,,$$
and
$$\begin{align*} (F_{n-1}+F_{n+1})-(2F_n-2)&=(F_{n-1}-F_n)+(F_{n+1}-F_n)+2\\ &=-F_{n-2}+F_{n-1}+2\\ &=F_{n-3}+2\\ &>0\,, \end{align*}$$
i.e., $F_{n-1}+F_{n+1}>\deg B_{n-2}(x)$, so $B_{n-2}(x)$ and $x^{F_{n-1}+F_{n+1}}C_{n-2}(x)$ have no powers of $x$ in common. Since $B_{n-2}(x)$ and $x^{F_{n-1}+F_{n+1}}C_{n-2}(x)$ are both timid, this implies that $A_n(x)$ is timid as well.
We can use a similar argument to show that $B_n(x)$ is timid:
$$\begin{align*} B_n(x)&=A_{n-2}(x)\left(1-x^{F_n}\right)\left(1-x^{F_{n+1}}-x^{F_{n+2}}\right)\\ &=A_{n-2}(x)\left(1-x^{F_{n+1}}\color{red}{-x^{F_{n+2}}}-x^{F_n}\color{red}{+x^{F_n+F_{n+1}}}+x^{F_n+F_{n+2}}\right)\\ &=A_{n-2}(x)\left(1-x^{F_n}-x^{F_{n+1}}\right)+x^{F_n+F_{n+2}}A_{n-2}(x)\\ &=B_{n-1}(x)+x^{F_n+F_{n+2}}A_{n-2}(x)\,, \end{align*}$$
where $\deg B_{n-1}(x)=2F_{n+1}-2<F_n+F_{n+2}$, so $B_{n-1}(x)$ and $x^{F_n+F_{n+2}}A_{n-2}(x)$ have no powers of $x$ in common. Since both are timid, so is their sum, $B_n(x)$.
Finally,
$$\begin{align*} C_n(x)&=A_{n-2}(x)\left(1-x^{F_n}\right)\left(1+x^{F_n}-x^{F_{n+2}}\right)\\ &=A_{n-2}(x)\left(1\color{red}{+x^{F_n}}-x^{\color{green}{F_{n+2}}}\color{red}{-x^{F_n}}-x^{2F_n}+x^{\color{brown}{F_n+F_{n+2}}}\right)\\ &=A_{n-2}(x)\left(1-x^{\color{green}{2F_n+F_{n-1}}}-x^{2F_n}+x^{\color{brown}{2F_n+F_{n+1}}}\right)\\ &=A_{n-2}(x)-x^{2F_n}A_{n-2}(x)\left(1+x^{F_{n-1}}-x^{F_{n+1}}\right)\\ &=A_{n-2}(x)-x^{2F_n}C_{n-1}(x)\,, \end{align*}$$
where $\deg A_{n-2}(x)=\sum\limits_{k=2}^{n-1}F_k=F_{n+1}-2$, and
$$\begin{align*} 2F_n-(F_{n+1}-2)&=F_n+(F_n-F_{n+1})+2\\ &=F_n-F_{n-1}+2\\ &=F_{n-2}+2\\ &>0\,. \end{align*}$$
Thus, $\deg A_{n-2}(x)<2F_n$, $A_{n-2}(x)$ and $x^{2F_n}C_{n-1}(x)$ are timid and have no powers of $x$ in common, so $C_n(x)$ is timid. This completes the induction, and we conclude that $A_n(x)$ is timid for each $n\ge 1$.
Note that we get for free the result that when
$$\begin{align*} \prod_{n\ge 2}\left(1-x^{F_n}\right)&=(1-x)\left(1-x^2\right)\left(1-x^3\right)\left(1-x^5\right)\left(1-x^8\right)\ldots\\ &=1-x-x^2+x^5+x^7-x^8+x^{11}-x^{12}-x^{13}+x^{14}+\ldots \end{align*}$$
is expanded as a formal power series $\sum\limits_{n\ge 0}a_nx^n$, $a_n\in\{-1,0,1\}$ for each $n\ge 0$: the coefficient $a_n$ of $x^n$ depends only on the factors $1-x^{F_k}$ with $F_k\le n$, so it is equal to the coefficient of $x^n$ in $A_k(x)$ for all $k$ such that $F_{k+1}\ge n$.
By the way, the coefficients $a_n$ have a nice combinatorial interpretation. Let $r_E(n)$ ($r_O(n)$, respectively) be the number of partitions of $n$ into an even (odd, respectively) number of distinct parts drawn from the set $\{F_k:k\ge 2\}$; then $a_n=r_E(n)-r_O(n)$, since each partition of $n$ into an even number of parts contributes a term $x^n$, and each partition into an odd number of parts contributes a term $-x^n$. (The coefficient of $x^n$ in $A_m(x)$ has a similar interpretation with the sizes of the parts limited to the set $\{F_2,\ldots,F_{m+1}\}$.)
Federico Ardila, The coefficients of a Fibonacci power series, Fibonacci Quarterly, $\mathbf{42}\,(2004)$, no. $3$, $202$-$204$, makes use of this interpretation to prove a recurrence that allows the $a_n$ to be computed quickly and easily in blocks of the form $[F_m\le n<F_{m+1}]$.
Two final observations:
The sequence $\langle 0,1,2,4,7,8,11,\ldots\rangle$ of exponents of non-zero terms of the power series $\sum_{n\ge 0}a_nx^n$ is OEIS A151661.
For $n\ge 1$ let $t_n$ be the number of non-zero terms in $A_n(x)$ then $t_n=2a(n)$, where $a(n)$ is given by OEIS A104767. (This follows from the comment there by Michael Somos.) Thus, $t_1=2$, $t_2=4$, $t_3=6$, and $t_n=2t_{n-1}-2t_{n-2}+2t_{n-3}$ for $n\ge 4$.