Take b = 2.

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

11 is prime. b^((n-1)/11)-1 mod n = 49, which is a unit, inverse 809.

5 is prime. b^((n-1)/5)-1 mod n = 798, which is a unit, inverse 267.

(5 * 11) divides n-1.

(5 * 11)^2 > n.

n is prime by Pocklington's theorem.