Take b = 2.

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

7 is prime. b^((n-1)/7)-1 mod n = 170, which is a unit, inverse 36.

5 is prime. b^((n-1)/5)-1 mod n = 106, which is a unit, inverse 2.

(5 * 7) divides n-1.

(5 * 7)^2 > n.

n is prime by Pocklington's theorem.