Let $p\in \Bbb{P},\alpha\in \Bbb{N}$. Find all of the solutions of the following equation: $$(p-1)!+1=p^{\alpha} \ \ \ ,p>6$$
My attempt
We can rewrite the equation as follows:
$$(p-1)!=p^{\alpha}-1$$ $$(p-1)!=(p-1)(1+p+p^2+...+p^{\alpha-1})$$ $$(p-2)!=1+p+p^2+...+p^{\alpha-1}$$
Then the best I could find out is that:
$$(p-2)!\equiv \alpha \pmod{p-1}$$
And in general if $ 2\leq k \leq p-1$:
$$k^{\alpha}=1 \pmod{p-k}$$
But then I don't know how to continue. Thank you for your time
Observe if $p>6$ then $(p-1)^2 | p^\alpha -1$ and hence $p-1 |\frac{p^{\alpha}-1}{p-1}=\sum_{i=0}^{\alpha-1} p^i\implies p-1 | \alpha \implies \alpha \geq p-1 $ but then $p^{\alpha}\ge p^{p-1}> (p-1)!+1$