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Let $k\in\mathbb{N},~k>5$ be a prime number. Prove $(k-1)!+1$ has at least two different prime divisors.
I am thinking on Wilson's Theorem, but I don't know how to use it.

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    @AnuragA This is not a duplicate, I need two **different** divisors. – 2017-01-11
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    The linked question answers yours though: since it cannot be a prime power, it has two different divisors. – 2017-01-11
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    @JoseA132 do you need to show these two different divisors? – 2017-01-11
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    @Xam I just have to prove it, not to show it. – 2017-01-11
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    @AnuragA No, I found this exercise exactly like I post it. – 2017-01-11
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    @SteveD The linked question shows $(k-1)!+1$ is not a power of $k$. It doesn't show $(k-1)!+1$ is not a power of any prime. – 2017-01-11
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    @yurnero actually it still does because Wilson's theorem guarantees that $k$ is a prime divisor so if we can show it is not a power of $k$, then it shows existence of two distinct prime divisors. – 2017-01-11

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By Wilson's theorem $k$ is a prime number $\iff (k-1)! \equiv -1 \pmod k$

Then $(k-1)!+1 \equiv 0 \pmod{k}$ this means that $k \mid (k-1)!+1$

Since $k>5$ and $k$ is proved to be prime then $(k-1)! + 1 = k.a$ where $k$ is one prime factor and $a$ is other prime factor or a composite number involving prime factors.

Note that $a = k$ when $\frac{(k-1)!+1}{k} = k$ and this is easily proven taking primes $k=\{2,3,5\}$ showing that $7$ the quotient of the aforementioned is $>k$. This give us the condition $k > 5$

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    But what if $a=k$? – 2017-01-11
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    You need to still prove $a$ is not a power of $k$. – 2017-01-11
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    $a \neq k$ when $\frac{(k-1)!+1}{k} \neq k$ and this is easily proven taking primes $k=\{2,3,5\}$ showing that in $7$ the quotient of the aforementioned is $>k$. – 2017-01-11
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    @kub0x then I recommend you to edit your answer. – 2017-01-11
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    @Xam: Hope it fits in now. Thanks – 2017-01-11
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    @kub0x I realized that actually $a$ doesn't need to be equal to $k$ but maybe it's a power of $k$. So I think you're solution isn't right yet. – 2017-01-11
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    Then for $k>2, a=k^{n-1}$ when $k^n \equiv 1 \pmod{(k-1)!}$ and $a\neq k^{n-1}$ when $k^n \not \equiv 1 \pmod{(k-1)!}$ where $n \mid \varphi((k-1)!)$ thus $k$ is a primitive root $\mod p$ when $a\neq k^{n-1}$ But I would say that it's incomplete. – 2017-01-11