Can someone clarify this for me please?

Question

Okay, so the proof of the fact that the value of (p-1)! mod p is always p-1 is below:

If p is prime, the numbers {1,2,....p-1} form a group under multiplication modulo p. So for each element a in this group, there is an element b such that ab is congruent to 1 mod p.

If a is congruent to b mod p, then a^2 is congruent to 1 (mod p). This means that a^2 - 1 = (a+1)(a-1) = 0 (mod p). Since p is prime, this means a is congruent to 1 or -1 mod p (which means a = 1 or a = p-1).

Okay, so my question is, how did they get this part:
"If a is congruent to b mod p, then a^2 is congruent to 1 (mod p)."

Answer 1

because a=1 and 1^2= 1

Answer 2

From the previous paragraph, ab = 1 (mod p), so if a = b (mod p), then a^2 = ab = 1 (mod p). Cute, huh?