How to prove that $a=z^{p}$ for some $z \in \mathbb{Z_{+}}$? 
Claim : 
If for a positive, composite integer $a$  and an odd prime $p$, such that $\gcd(a,p)=1$, we are given   
$$  a^{p^{n-2}(p-1)} \equiv 1 \pmod {p^n} \ \forall \  n \geq 2 \ \ ;\  n \in \mathbb{Z_{+}} $$    
Then $a=z^{p}$ for some $z \in \mathbb {Z_{+}}$ 
Also note that the congruence holds for for all positive integers $n$, not just an arbitrary one.


My Approach :
At first a glance, it seems trivial to me, because by Euler's Theorem, $a^{\phi(p^{n})} \equiv 1 \pmod {p^n}$ and since $\phi(p^n) = p^{n-1}(p-1)$, I think that $a$ should be a perfect $p^{th}$ power to account for the extra $p$ in the power. 
But, I don't have a formal proof for this. Is my reasoning enough or do I need a more rigorous argument ?
Added Later : 
On observing more carefully, I found the following.  
Let the prime factorization of $a$ is $ \displaystyle \prod_{i=1}^{m} {q_{i}}^{k_{i}}$, where $q_{i}$ are prime factors of $a$. If we can prove that whenever the modular equation holds true, there exists at least one $k_{i} \geq p$, then we are done. This is because we can then write $a=j h^{p}$ and plugging it in the modular equation, we will get the same condition for $j$. This will set up an infinite descent and prove that $j=1$ and hence the claim will hold true.  
Query : As Mr. Greg Martin stated, the claim is false in general. However,  
If the modular equation is true for a fixed $n$, then,  
$a \equiv z^p \pmod {p^{n}}$  
$\displaystyle \therefore a \equiv {k_{1}}^p \pmod {p}$  
$a \equiv {k_{2}}^p \pmod {p^{2}}$
$.$
$.$
$.$
$a \equiv {k_{n}}^{p} \pmod{p^n}$
$\implies a = {k_{n}}^{p} + j p^{n}$  
If we choose $n$ to be sufficiently large, then $j=0$   
$\implies a={k_{n}}^{p}$  
Where is the fault in my reasoning? Please Help.
 A: The problem is false as stated.
Let $a$ be any integer such that $a^{p-1}\equiv1\pmod{p^2}$. Then
$$
a^{p(p-1)-1} = (a^{p-1}-1)\big( (a^{p-1})^{p-1} + (a^{p-1})^{p-2} + \cdots + (a^{p-1})^1 + 1 \big);
$$
the first factor is divisible by $p^2$ by assumption, and the second factor is congruent to $p\pmod {p^2}$, hence is divisible by $p$. We conclude that $a^{p(p-1)}\equiv1\pmod{p^3}$ automatically.
Similarly,
$$
a^{p^2(p-1)-1} = (a^{p(p-1)}-1)\big( (a^{p(p-1)})^{p-1} + (a^{p(p-1)})^{p-2} + \cdots + (a^{p(p-1)})^1 + 1 \big)
$$
is then divisible by $p^4$, etc. In short, one can prove by induction that if $a^{p-1}\equiv1\pmod{p^2}$, then automatically $a^{p^{n-2}(p-1)}\equiv1\pmod{p^n}$ for every $n\ge2$.
And $a^{p-1}\equiv1\pmod{p^2}$ certainly does not imply that $a$ must be the $p$th power of an integer. (Examine the pairs $(a,p) = (7,5)$ and $(a,p)=(17,3)$, for example.) And note that adding $p^2$ to $a$ preserves the congruence, so $(32,5)$ and $(57,5)$ and $(82,5)$ ... are all counterexamples as well; in particular, there are plentiful counterexamples where $a$ is not prime.
