TITLE: Using Bézout's Identity to prove that given $\gcd$ of two numbers is really true
QUESTION [5 upvotes]: I need to prove that
$$
\gcd(n!+1,(n+1)!+1) = 1
$$
In class, the teacher advised us to use Bézout's Identity. So I wrote the following
$$
x(n!+1) + y[(n+1)! + 1] = 1
$$
But from there I'm at a loss. 
To make the question more generic: I don't understand how Bézout's Identity can help me prove that the $\gcd$ of two numbers is really $n$. 
Previous exercises I had to prove were much simpler; for example, I had to prove that $\gcd(2n+1, 3n+1) = 1$. In this case, I just tried to find $x$ and $y$ that make the statement $x(2n+1) + y(3n+1) = 1$ true. And by finding any two integers, I understood that I had provided enough proof. But I didn't do any math to find $x$ and $y$ in that simpler case... or I did, but I don't understand what I did... anyway, I could just "see", right away, that if $x=3$ and $y=-2$, then it would work out.
Can you help me understand how to use Bézout's Identity in proofs of any type? I think I'm not using it correctly.

REPLY [4 votes]: Hint:

$(n+1)\cdot(n!+1)-\left[(n+1)!+1\right]=\color{red}n$
$(n!+1)-(n-1)!\cdot \color{red}{n}=1$

REPLY [3 votes]: I would not use "Bézout" directly. Let $d$ be a divisor of both. Then $d$ divides $(n+1)(n!+1)$ and $(n+1)!+1$. So it divides their difference, which is $n$. 
But if $d$ divides $n!+1$ and $n$, it divides $1$. 
Now if you wish you can work backwards, and find the appropriate multipliers. But we don't need them to prove relative primality.
Remark: In a sense, what we did is much like the procedure you described. You used $x$ and $y$ to get rid of the ugly part. We did the same thing in the first step, getting rid of the factorial stuff.