Relating two forms of little Fermat: $a^p\equiv a$ vs. $a^{p-1}\equiv 1$ if $p\nmid a$ I know that if $p$ prime and $p\nmid a$, then $a^{p-1}\equiv 1\pmod p$ and I know also that $a^{p}\equiv a \pmod p$ using the fact $a\equiv a \pmod p$ and multiplying the members.
What I couldn't understand is why in the Fermat little theorem we have  $a^{p}\equiv a \pmod p$ for all integer $a$.
 A: If $p$ divides $a$ then both $a^p$ and $a$ are congruent to $0$, so $a^p\equiv a\pmod{p}$ holds trivially. 
Remark: From $a^p\equiv a\pmod{p}$, we can conversely derive the more common form of Fermat's Theorem. For if $a^p\equiv a\pmod p$, then $p$ divides $a(a^{p-1}-1)$. So if $p$ does not divide $a$, then $p$ divides $a^{p-1}-1$. 
A: Either $p$ divides $a$ or $p\nmid a$
As $p$ is prime  $p\nmid a\implies (a,p)=1$
As $\displaystyle a^p-a=a(a^{p-1}-1)$
This will be divisible by $p$ if $p|a$
otherwise it will also be  divisible by $p$  as $p|(a^{p-1}-1)$ by Fermat little theorem
A: Following the comments of André Nicholas:
If $p\mid a$, then $a\equiv 0 \pmod p$, then multiplying $p$ times both sides we have $a^p\equiv 0 \pmod p$. Thus we have $a\equiv 0 \pmod p$ and $0\equiv a^p\pmod p$ by symmetry, then finally by transitivity: 
$$a\equiv a^p \pmod p$$
A: It is instructive to view this from a slightly more general perspective. The equivalence of the two common forms of the Little Fermat theorem is  special case $\, f(x) = x^{p-1}-1\,$ below.
Theorem $\ $  If  $\,p\,$ is prime and $f(x)$ is polynomial with integer coefficients then
$$\begin{align} 
      &p\mid x f(x)\ \text{ for all integers } x\\[.2em]
\iff\  &p\mid f(x) \ \ \ \:\! \text{ for all integers } x\ \text{coprime to } p
\end{align}$$
Proof $\ (\Rightarrow)\ $ By hypothesis  $\,p\mid xf(x),\,$ so $\,x\,$ coprime to $\,p\Rightarrow\,p\mid f(x)\,$ by Euclid's Lemma.
$(\Leftarrow)\ $ We split into two cases depending on whether or not $\,p\mid x$.
Case $(1)\ \ p\mid x.\ $ Then $\,p\mid xf(x).$
Case $(2)\ \ p\nmid x,\,$ so $\,x\,$ is coprime to $\,p,\,$ so by hypothesis $\,p\mid f(x)\,$ so $\,p\mid xf(x).\ $ QED
Remark $\, $ If you know modular arithmetic then it can be viewed more arithmetically as the following in the $\color{#0a0}{\rm domain}$ $R=\Bbb Z_p$ (true for a polynomials over any field or domain)
$$\begin{align} \forall x\!:&\ \ \ \ \ \ \ xf(x)= 0\\[.2em]
\iff\ \forall x\ &\ [\:\!x\not= 0\Rightarrow  f(x)= 0\:\!]\\[.6em]
{\rm generally}\quad \forall x\!:&\ \ \ \ \ \ g(x)\:\!f(x)= 0\\[.2em]
\iff \ \forall x\ &\  [\:\!g(x)= 0\ \ {\rm or}\ \ f(x)= 0\:\!]\ \ \text{by $\,R\,\ {\rm a}\ \color{#0a0}{\rm domain}$}\\[.2em]
\iff\ \forall x\ &\  [\:\!g(x)\not= 0\,\Rightarrow\:\!  f(x)= 0\:\!]\end{align}$$
