Why the $GCD$ of any two consecutive Fibonacci numbers is $1$? Note: I've noticed that this answer was given in another question, but I merely want to know if the way I'm using could also give me a proof.
I did the following:
$$F_n=F_{n-1}+F_{n-2} \\ F_n=[F_{n-2}+F_{n-3}]+[F_{n-3}+F_{n-4}]\\F_n=F_{n-2}+2F_{n-3}+F_{n-4}\\ F_n=[F_{n-3}+F_{n-4}]+2[F_{n-4}+F_{n-5}]+[F_{n-5}+F_{n-6}]\\F_n=F_{n-3}+3F_{n-4}+3F_{n-5}+F_{n-6}\\ \dots \tag{1}$$
I guess that the coefficients of the $F_n$'s might indicate something that could prove it. But I'm not sure if it's possible. Perhaps the impossibility of writing the expression as:
$$n(b_1a_1+b_2a_2+\dots +b_n a_n)$$
With $n,b_n\in\mathbb{N}$ would show that. But I'm not sure on how to proceed. This should be true because if $a$ and $b$ have a common divisor $d$, then:
$$a+b=a'd+b'd=d(a'+b')$$
It is possible to extend this:
$$a+b+c=da'+db'+dc'=d(a'+b'+c')$$
I have noticed that the numbers that appear in the expansion I've shown in $(1)$ seems to be the Pascal's triangle. So perhaps these numbers as coefficients of the $F_n$'s might indicate that it's not possible to write them as: 
$$d(a'+b'+c')$$

If $\,f_n = a f_{n-1}\! + f_{n-2}\,$ then induction shows $\,(f_n,f_{n-1}) = (f_1,f_0)\,$ since $\, (f_n,f_{n-1}) = (a f_{n-1}\! + f_{n-2},\,f_{n-1}) = (f_{n-2},f_{n-1}) = (f_1,f_0)\,$ by induction
Remark $\ $ Similarly one can prove much more generally that the Fibonacci numbers  $\,f_n\:$ comprise a strong divisibility sequence: $\,(f_m,f_n) = f_{(m,n)},\:$ i.e. $\,\gcd(f_m,f_n) = f_{\gcd(m,n)}.\:$ Then the above  is just the special case $\,m=n\!+\!1.\:$
