**Hint** $\ $ Put $\rm\:a_n = 1 = b_n\:$ in the much more general result below.

**Theorem** $\ $ If $\rm\:(\color{#c00}{b_n,\,f_n) = 1}\:$ and $\rm\:f_{n+1} = a_n f_n + b_n f_{n-1}\:$ then $\rm\:(f_{n+1},\,f_n) = (f_1,\,f_0).$

**Proof** $\ $ Clear if $\rm\:n = 0.\:$ Else by Euclid and induction we have

$$\rm (f_{n+1},\,f_n) = (a_n f_n\! +\! \color{}{b_n} f_{n-1},\,f_n) = (\color{#c00}{b_n} f_{n-1},\, \color{#c00}{f_n}) = (f_{n-1},\,f_n) = (f_1,\,f_0)\qquad$$

**Remark** $ $ Similarly we can prove much more generally that the Fibonacci numbers $\rm\:f_n\:$ comprise a *strong* divisibility sequence: $\rm\, (f_m,f_n) = f_{(m,n)},\:$ i.e. $\rm\:gcd(f_m,f_n) = f_{\gcd(m,n)}.$ OP is case $\rm\,m=n\!+\!1.$