Proof
(related to Proposition: A Necessary Condition for an Integer to be Prime)
 According to the explicit formula for the Euler function, we have $\phi(p)=p1$ for any prime number $p.$
 Note that if $p$ is not a divisor of $a$ then both are coprime.
 Therefore, $a^{p1}(p)\equiv 1(p)$ follows immediately from Fermat's little theorem for a prime module.
 Multiplying both sides gives us the congruence with $a^p(p)\equiv a(p).$
 If $p\mid a,$ then trivially $a^p(p)\equiv 0(p)\equiv a(p).$
 Therefore, the congruence $a^p(p)\equiv a(p)$ holds for all prime numbers $p$ and all integers $a,$ (i.e. no matter if $p\mid a$ or $p\not\mid a$).
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927