Non-constructive proof for a prime $p, \forall k \in \mathbb{Z+}, a\equiv b\pmod {p^k}\implies a\equiv b\pmod p.$

Question

As a simple case take $k=2, a\equiv b\pmod {p^2}\implies a\equiv b\pmod p.$
A constructive example for $a= 2, b=27, p =5$ will work, but needed is a general or non-constructive approach, so as to get the reasoning applicable generally in a formal manner.

I mean the reason for the no change in residue class, for a root of the modulus is not clear.

I can only work out the case when $a , b \lt p$, as it is the trivial case (with $a=b$) for the smaller modulus case.

Answer 1

$$a \equiv b \pmod{p^k}$$

$$a-b \equiv 0 \pmod{p^k}$$

$$\exists m \in \mathbb{Z}, (a-b) =p^km$$

$$\exists m \in \mathbb{Z}, (a-b) =p(p^{k-1}m)$$

$$a-b \equiv 0 \pmod{p}$$

Answer 2

If $a|b $ and $b|c $ then $a|c $. (Because if there exist $k $ so that $b=a*k $ and if there is a $m $ so that $c=b*m$, then $c=a (k*m) $

So if $a\equiv b\mod p^2$ that means $p^2|a-b $. And since $p|p^2$ we know $p|a-b $