Show that $p!$ and $(p - 1)! - 1$ are relatively prime

Question

If $p$ is prime number, with $p>3$

Show that $p!$ and $(p - 1)! - 1$ are relatively prime.

I tried

$\text{gcd}\;(p!,(p-1)!-1)=d\Longrightarrow d\mid p!$ e $d\mid(p-1)!-1$ having to $p!=p\cdot(p-1)!$ and $d\mid p!$ then, $$d\mid p\cdot(p-1)!$$

Soon $d\mid p$ or $d\nmid p$.

(I)

If $d\mid p\Longrightarrow d=1$ or $d=p$

We have $d\mid (p-1)!-1$ and Wilson's Theorem we have $$p\mid (p-1)!+1$$and $(p-1)!-1=(p-1)!+1-2$ We know by Theorem Wilson $p\mid (p-1)!+1$, then $$p\nmid (p-1)!+1-2$$ because $p>3$, then $p\neq d$, soon $d=1$.

(II)

If $d\nmid p\Longrightarrow d\mid (p-1)!$

As $d\mid (p-1)!$ and $d\mid (p-1)!-1$, then $d\mid 1\Longrightarrow d=1$.

$\Box$

You demonstration is complete, or is something missing? Something wrong? Another way?

Answer 1

Since $(p-1)!-1$ is relatively prime to every integer smaller than $p,$ the only way the two quantities could fail to be relatively prime is if $p|(p-1)!-1.$ But $p|(p-1)!+1,$ and $p>2,$ so that's impossible.

Answer 2

Your proof is nearly complete. For part (II), I would suggest that instead of concluding that $d=1$, conclude that the assumption for this part (that $d\nmid p$) contradicts the result that $d\mid 1$, therefore that this case leads to a contradiction and cannot occur.

Then conclude that the result from part (I) is the only logical result, which is that

$$\text{gcd}(p!,(p-1)!-1)=1$$