How to show that $p!+1\equiv 1 \mod k$ I am a non mathematician who is taking a self study class in number theory. I was wondering how to formally prove the following:
Let $p$ be a prime number. How can I show that $$p!+1\equiv 1 \mod k$$ for any integer $2\le k\le p.$
 A: What you have written is true not just for prime $p$ but for all $p$ as well. All you need to show is that $p! \equiv 0 \pmod{k}$ for $p \in \{1,2,\ldots,n\}$. Write out $p!$ as
$$p! = 1 \times 2 \times 3 \times \cdots \times p$$
and now...
A: if $2\leq k\leq p$ then by definition $k$ is in the product $1\cdot2\dots p$. This means $k|p!$ implying $p!=kn$ for some $n\in\mathbb{Z}$. This implies $p!+1=kn+1$. Therefore what you want to show follows. 
A: You can take away 1 from both sides and not affect the congruence.
$$p! \equiv 0 \mod k$$
Note that 
$$p!=p \cdot (p-1)\cdot(p-2)\cdot...\cdot3\cdot2\cdot1$$
So $p!$ is a multiple of any $2 \ge k \ge p$, so $p!$ leaves a remainder of $0$ when divided by $k$.
A: It's clear that $k$ divides $p!= 1 \times 2 \times 3 \times \cdots \times p$ if $2\leq k\leq p$ so $p!\equiv 0 \mod k$ and hence $p!+1\equiv 1 \mod k$
A: $$p! + 1 \equiv 1 \pmod k \iff p! + 1 - 1 \equiv 1 - 1 \pmod k \iff p!  \equiv 0 \pmod k$$
$$p! \equiv 0 \pmod k \iff k \mid p\,!\tag{1}$$
Since $2 \leq k \leq p$, if follows that $k$ is one of the factors in $p!$: $$p\,! = p\times (p - 1) \times \cdots \times k \times \cdots 2 \times 1$$
That is, as a factor of $p!$, $k$ divides $p!$: $\;\;k \mid p!$.
Hence by $(1),\;$ $p! \equiv 0 \pmod k \iff p! + 1 \equiv 1 \pmod k$
