Prove $(n!-1,(n-1)!-1)=1$

Question

Question: Let $n\geq2,n\in\mathbb{N}$. Prove $(n!-1,(n-1)!-1)=1$

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I have noticed that $n!=n\cdot (n-1)!$

So letting $\alpha=(n-1)!$, we have to prove $(n\alpha-1,\alpha-1)=1$

I feel that this is just a simple relationship, and it seems intuitively obvious that this is the case, but I cant prove it.

Where can I find a proof for this relationship? Any help would be greatly appreciated.

Answer 1

If integer $d$ divides both, $d$ must divide $n!-(n-1)!=(n-1)!\cdot(n-1)$

But $((n-1)!\cdot(n-1),(n-1)!-1)=1\implies ((n-1)!\cdot(n-1),d)=1$

Answer 2

You can just apply the Euclidean algorithm here. Since $n\times((n-1)!-1)=n!-n$ you get $$\begin{align} \gcd(n!-1,(n-1)!-1)&=\gcd((n!-1)-(n!-n),(n-1)!-1)\\ &=\gcd(n-1,(n-1)!-1)\\ &=\gcd(n-1,-1)\\&=1 \end{align} $$ since $n-1$ obviously divides $(n-1)!$ for $n>1$. By the way you see why $n\geq2$ was given: for $n=1$ the latter step does not apply, and in fact one gets $\gcd(n!-1,(n-1)!-1)=\gcd(0,0)$ which depending on your definitions is either undefined or defined to be$~0$.

Answer 3

Lemma $\,\ (\overbrace{n(n\!-\!1)\, k-1}^{\Large\color{#c00} a},\ \overbrace{(n\!-\!1)\, k-1}^{\Large\color{#0a0} b})\ =\ 1\ \ \,$ [yours is $\ k = (n\!-\!2)!\:$]

Proof $\ \,$ Modulo the $ $ gcd $\,d = (\color{}a,\color{}b)\,$ we have

$\begin{array}{lrl} (1)&\quad\ n(n\!-\!1)\, k\ \equiv&\!\!\! 1 & {\rm by\ }\ d\mid\color{#c00} a\,\Rightarrow\,a\equiv 0\,\Rightarrow\,a\!+\!1\equiv 1\\ (2)& (n\!-\!1)\, k\ \equiv&\!\!\! 1 & {\rm by\ }\ d\mid\color{#0a0} b\ \Rightarrow\,b\equiv 0\,\Rightarrow\,b\!+\!1\equiv 1 \\ (3)& n\ \equiv&\!\!\! 1 & {\rm by\ substituting}\ (2)\ {\rm in}\ (1)\\ (4)& 0\ \equiv&\!\!\! 1& {\rm by\ substituting}\ (3)\ {\rm in}\ (2)\\ \end{array}$

This $ $ implies $\ d\,\mid\, 0\, -\, 1,\ $ hence $\,\ d = 1\quad$ QED