Take b = 2.

b^(n-1) mod n = 1.

7 is prime. b^((n-1)/7)-1 mod n = 63, which is a unit, inverse 107.

3 is prime. b^((n-1)/3)-1 mod n = 127, which is a unit, inverse 69.

(3 * 7) divides n-1.

(3 * 7)^2 > n.

n is prime by Pocklington's theorem.