If $a\equiv b [p^k]$ then $a^p \equiv b^p [p^{k+1}]$

Question

Can anyone explain the steps to this proof? I'm really lost/

If $k\geq 1$ and $a\equiv b[p^k]$ then $a^p \equiv b^p [p^{k+1}]$

Proof:

Since $a= b + qp^k$ for some $q\in \mathbb{Z}$ we have $a^p = (b+qp^k )^p = b^p + p^{b^{p-1}} \cdot qp^k + mp^{2k} =b^p + (b^{p-1}q + mp^{k-1})p^{k+1}$

How do you get from $(b+qp^k )^p $ to $ b^p + p^{b^{p-1}} \cdot qp^k + mp^{2k} $

Answer 1

$$(b+qp^k)^p=b^p+\binom{p}{1}b^{p-1}(qp^k)^1+\binom{p}{2}b^{p-2}(qp^k)^2+\cdots$$ All terms after the first two have at least $p^{2k}$ in them, so $$(b+qp^k)^p=b^p+pb^{p-1}qp^k+\text{stuff}\times p^{2k}=b^p+p^{k+1}(b^{p-1}q+\text{stuff}\cdot p^{k-1})$$ Where what I have called "stuff" they called $m$.

Answer 2

Just a hint : $ a^{p} = \Big(b+ q p^{k}\Big)^p=\sum_{i=0}^p{p\choose i}\Big(qp^{k}\Big)^i \Big(b\Big)^{p-i}$

Then notice that for $i\ge 2$ you have $ik\ge k+1$, so $\sum_{i=2}^p{p\choose i}\Big(qp^{k}\Big)^i \Big(b\Big)^{p-i} \equiv 0 \ [p^{k+1}]$

That is why $a^p = b^p + qp^{k+1}b^{p-1} \equiv b^p \ [p^{k+1}]$