Problem regarding Euler's Theorem: $a^{\phi(n)}\equiv 1 \bmod n$ Here's the problem:

If $n \geq 2$, and if $p$ is a prime number s.t $p|n$ but $p^2$ is not a factor of $n$, then $$p^{\phi(n)+1}\equiv p \mod n$$

So since we're dealing with Euler's Phi function, I figured that this was an application of Euler's Theorem (please correct me if I am wrong). An attempt (though miniscule):

Attempted Proof
We know that $p|n$, so $\exists q \in \mathbb{Z}$ s.t
$$n=pq.$$
However, since $p^2$ is not a factor of n, we know that these integers are relatively prime to eachother. Thus, $\exists r,t \in \mathbb{Z}$ s.t
$$1=p^2r+nt$$
Now if we consider the integers modulo n, we see that $p^2$ is invertible since it is relatively prime to $n$ and that $r$ happens to be our inverse. Thus, we can say that
$$1=p^2r + nt$$
$$\implies 1-p^2r=nt$$
$$\implies [1-p^2r]_n=[0]_n$$
$$\implies [p^2]_n[r]_n=[1]_n$$
So $r$ must be the inverse of $p^2$.

Okay so I'm sure I've gone off into a tangled direction. So my question is: how am I to finally get to the $\phi$ function? I mean, considering how Euler's Theorem goes, we know that we have
$$a^{\phi(n)} \equiv 1 \mod n$$
So wouldn't it follow naturally that
$$a^{\phi(n)+1}\equiv a \mod n$$
????
For our problem, we of course have $a=p$.

I think I've got it now:
We know that $p|n$ $\implies$ $\exists \gamma \in \mathbb{Z^{+}}$ s.t $n=p\gamma$. We are given the $p^2$ is not a factor of $n$ $\implies p^2 > n$ and so $p$,$\gamma$ share no common divisors. Thus, $\gcd(p,\gamma)=1$. Then by Euler's Theorem, we have
$$p^{\phi(\gamma)} \equiv 1 \mod p$$
Since $\phi(n)=\phi(p)\phi(\gamma)$, it follows that
$$p^{\phi(p) \cdot \phi(\gamma)} \equiv 1 \mod p \implies (p^{\phi(\gamma)})^{\cdot \phi(p)} \equiv 1 \mod \gamma$$
$$\implies p^{\phi(n)} \equiv 1 \mod \gamma$$
So it follows that $\gamma | (p^{\phi(n)} - 1)$ $\implies$ $\gamma \cdot p |(p^{\phi(n)+1}-p)$ $\implies$ $n|(p^{\phi(n)+1}-p)$. Hence
$$p^{\phi(n)+1} \equiv p \mod n$$ 
