Confirmation of Proof: $p\mid n \implies p\nmid n+1$ So, I am having trouble approaching this question, for a direct proof.

Prove that:
For any integer $n$ $($expressed as $n \in \mathbb{Z})$ and prime number $p$,
if $p\mid n \Rightarrow p\nmid n + 1$

(I know it is considered trivial, and is taken as a given, but,)
I would appreciate some (preferably short, and) elegant proof for the statement.
Note: Proof by contraposition, contradiction, exhaustion (cases), etc. all ideas and contributions are welcome. (With all the different ideas, it would be possible to attempt to combine them to come up with a more compact proof).
 A: $$p|n \implies n=pk \implies n+1=pk+1$$
Thus dividing $n+1$ by $p$ results with a remainder of $1$.
A: Well, taking the contrapositive gives you the statement "if $p|(n+1)$ then $p\nmid n$", which isn't terribly helpful since it's very similar to the original statement. And proof by exhaustion doesn't seem terribly promising either, since there are infinite choices for $n$ and $p$ and no clear way of grouping them. So let's try contradiction:
Assume $p$ is prime, $n$ is an integer, and $p|n$. Assume towards contradiction that $p|n+1$. Then by definition of divisibility, there exist integers $a, b$ such that $ap = n$ and $bp = n+1$.
Can you use this to get a contradiction? (Hint: since $p$ is prime, $p \geq 2$.)
A: Note that $\mbox{gcd}(n,n+1)=1.$ Supposing by contradiction that $p\vert n+1,$ then $p$ is a common divisor of $n$ and $n+1$ and hence $p\vert 1$. Therefore, $p$ must be equal to 1, which is a contradiction.
A: "I know it is considered trivial"
So let's think why.
If $p|n$ means $n = kp$ is a multiple of $p$.  Well, what is the very next possible multiple of $p$?  It would have to be $p(k+1) = kp + p = n + p> n+1$.
$n+1$ is too small to be a multiple of $p$ if $n$ is a multiple of $p$, because to get the next multiple of $p$ we must add $p$ and $p$ is bigger than $1$.
.......
There are, of course theorems, that state things like $\gcd(n, n+1) = \gcd(n, (n+1) -n) = \gcd(n, 1) =1$ so $n$ and $n+1$ are relatively prime. Or if $p|n$ and $p|m$ then $p|n-m$.  So if $p|n$ and $p|n+1$ then $p|(n+1) -n = 1$ and $p|1$ is a contradiction, etc.
Those are all true and they all rely on the same concept (that the differences of multiples are multiples).  But I want it to be clear that the multiples of $p$ are each a "jump" of $p$ away.
......
Oh..... here's another obvious way.
If $n = pk$ then $n+1 = pk + 1 = p(k + \frac 1p)$ and $k + \frac 1p$ is not an integer.
......  
Again, the basic idea is that the multples of $p$ are: $p, 2p, 3p, 4p,..... kp, (k+1)p,....$.  There's simply "no room" to fit $mp, mp + 1, mp+p$.
A: Correct me if wrong:
Let $r,p,n \in \mathbb {N^+}$.
1)Given : $p|n \rightarrow n=rp.$
2) Assume: $p|(n+1)$  $\rightarrow$ $n+1 = sp$.
Subtract equation 1) from equation 2):
$1= (s-r)p$  $\rightarrow$  $p|1.$
A prime $2,3,..$. divides $1$?
