Looking to understand part of this lemma related to primitive roots

Question

Hi I am having trouble following part of a proof of a small lemma and am looking to see if anyone can help explain it

The lemma is that if $P$ is an odd prime and if $a=b+P^{k}$ with gcd(P,bm)=1 then $a^{P}=b^{P}+P^{k+1}M$ with gcd(M,P)=1

The proof writes that $a^{P}=b^{P}+P^{k+1}mb^{P-1}modP^{k+2}$

$a^{P}=b^{P}+P^{k+1}mb^{P-1}+cP^{k+2}$

How do they get this ? How does the $P^{k+2}$ come into this ?

I know the binomial expansion for exponent p ie $(a+b)^{P}=a^{P}+b^{P}$ mod p but that does not seem to be used .

Any help explaining this ? Thanks

Answer 1

HINT: $$X \equiv Y \pmod {p} \iff p \mid X - Y \iff X - Y = pc \iff X = Y + pc$$