I have to proof $gcd(F\_{n+1},F\_{n+2})$ need at least n steps to get gcd(we have to use version with subtraction).

My try

After first step we get $gcd(F\_{n+1} - F\_{n+2},F\_{n+2})=gcd(F\_{n+1},F\_{n})$

After second step we get $gcd(F\_{n+1}-F\_{n},F\_{n})=gcd(F\_{n-1},F\_{n})$

After n-th step we get $gcd(F\_{3}-F\_{2},F\_{2})=gcd(F\_{1},F\_{2})$

So after n+1-th step we get $gcd(F\_{1},F\_{2}-F\_{1})=gcd(F\_{1},F\_{0})$

That show us we need n+1 steps to get gcd of 2 fibonnaci numbers.

But it isn't formaly (I'm not sure that's proof) could someone help me how that should looks correctly