The coefficients of the power series $\frac{1}{1-z-z^2}$ centered at $0$ are the numbers in the Fibonacci sequence So I need to show the coefficients of the power series centered about $z=0$ of $\frac{1}{1-z-z^2}$ are $1,1,2,3,5,8,13,21,\dots$
Here's what I have thus far but I am unable to show all coefficients are equal to the coefficients of the series.
Let $F(z):= \sum_{n=0}^\infty a_nz^n$
Then
\begin{align}
\sum_{n=0}^\infty a_nz^n &= a_0 + a_1z + z^2 \sum_{n=0}^\infty (a_{n+2})z^n \\
&= 1+z+ z^2 \sum_{n=0}^\infty (a_{n+1}+a_n)z^n\\
&= 1+ z+ z \sum_{n=0}^\infty a_nz^n - 1 + z^2(\sum_{n=0}^\infty a_nz^n)\\
&= 1+ z(\sum_{n=0}^\infty a_nz^n)+z^2(\sum_{n=0}^\infty a_nz^n)
\end{align}
Then setting $F(z)= \sum_{n=0}^\infty a_nz^n$, bringing everything to the other side and dividing we obtain
$$F(z) = \frac{1}{1-z-z^2}$$
But I am stuck here. I saw another version of this asked but it uses the Cauchy Integral formula which we have not yet covered in the course. The hint was to write $\sum_{n=0}^\infty a_nz^n$ and use the recursive formula to rewrite the power series.

One more way:
Let $a,b=\frac{1\pm\sqrt{5}}{2},a-b=\sqrt{5}$, we know that Fibonacci numbers are $f_n=\frac{a^n-b^n}{\sqrt{5}}.$
$$F(x)=\frac{1}{1-x-x^2}=\frac{1}{(1-ax)(1-bx)}=\frac{1}{x\sqrt{5}}\left[\frac{1}{1-ax}-\frac{1}{1-bx}\right].$$
Use IGP: $\frac{1}{1-z}=1+z+z^2+z^3+....., |z|<1$, then
$$F(x)=\sum_{k=0}^{\infty}\frac{(a^k-b^k)x^k}{\sqrt{5}}=\sum_{k=0}^{\infty}f_k ~x^k, \quad |x| <\frac{\sqrt{5}-1}{2}. $$
