Two congruence. Are they true?

Question

Let $p$ be prime, $n \in \mathbb{N}$, $l \in \mathbb{N}$

The first one: $$ n \neq pl \implies \forall k\in \mathbb{N} \ \ n^{\phi (p^k)} \equiv 1 \pmod p. \\ $$ And the second one: $$ (1+n)^p = 1+ n \mod p. $$

Are they true?

Answer 1

As $\varphi(p^k)=p^{k-1}(p-1)$ and $n^{p-1}\equiv 1 \mod p$ for all $\not\equiv 0 \mod p$ (Little Fermat), we have $$n^{\varphi(p^k)}=\bigl(n^{p-1}\bigr)^{p^{k-1}}\equiv 1 \mod p.$$

For the second congruence, it is even more general: $$(n+n')^p\equiv n^p+n'^p \mod p$$ because one shows that, if $p$ is prime, all binomial coefficients $\displaystyle\binom pk$ are divisible by $p\,$ if $\,0<k<p$. So $$(1+n)^p\equiv 1+n^p \equiv 1+n $$ by Little Fermat again : $\,n^{p-1}\equiv 1 \mod p\Rightarrow n^p\equiv n$ for all $n\not\equiv 0 \mod p$, and $n^p\equiv n\,$ also if $n\equiv 0\mod p$.