Sorry, don't know a reference, but here is a quick argument.

If $M=p^ab+c$ with $0\leq b\leq p^a-1$, then $$(1+x)^M=(1+x)^{p^ab}(1+x)^c =(1+x^{p^a})^b(1+x)^c \mod p. $$ In turn, this equals $$ (1+bx^{p^a} + ...)(1+x)^c \mod p $$ where $...$ means higher degree terms. Since $c<p^a$, there is no further correction after multiplying by $(1+x)^c$ (the coefficient by $x^{p^a}$ stays $b$), and you are done.

Sorry, don't know a reference, but here is a quick argument.

If $M=p^ab+c$ with $0\leq c\leq p^a-1$, then $$(1+x)^M=(1+x)^{p^ab}(1+x)^c =(1+x^{p^a})^b(1+x)^c \mod p. $$ In turn, this equals $$ (1+bx^{p^a} + ...)(1+x)^c \mod p $$ where $...$ means higher degree terms. Since $c<p^a$, there is no further correction after multiplying by $(1+x)^c$ (the coefficient by $x^{p^a}$ stays $b$), and you are done.