\[p\]

, \[x^{p-1}=1 \equiv (x-1)(x-2)...(x-(p-1)) \; (mod \; p) \]

.It is a consequence of Lagrange's Theorem that a polynomial of degree

\[k\]

has at most \[k\]

possible roots. Any polynomial of the form above has exactly \[p-1\]

roots \[(mod \; p)\]

.This is a direct consequence of Fermat's Little Theorem since the equation

\[x^{p-1} -1 \equiv 0 \; (mod \; p)\]

is satisfied by the \[p-1\]

incongruent number \[x \equiv 1, \; 2, \; 3,..., \; p-1\]

. Hence, by Factorising a Congruence\[x^{p-1}=1 \equiv (x-1)(x-2)...(x-(p-1)) \; (mod \; p) \]

.