Common...but interesting enough

If a₁, a₂,..., a_φ(m) are the numbers less than m and prime to it.

Consider the number a with (a,m) = 1

Now. the numbers aa₁, aa₂,..., aa_φ(m) are again all prime to m. Further they are all distinct modulo m as for all r, s ≤ φ(m) we have a (a_r-a_s) is not divisible by m.

That means by pigeon-hole aa₁, aa₂,..., aa_φ(m) are congruent to a₁, a₂,..., a_φ(m) in some order

Hence ( aa₁) (aa₂) ... (aa_φ(m)) ≡ a₁ a₂ a_φ(m) mod m

or a^φ(m)≡1 mod m

thanx....