Proof that if the set $F_p={0, 1, ..., p-1}$ is a field defined with addition and multiplication mod p, then $p$ must be a prime

Question

I'm studying about the field of Galois (Finite Fields) and I'm going through a proof that tries to show that if a finite set $F_p$ is a field, then $p$ must be a prime.
*note: I cannot link the materials since they're in foreign language

I don't understand the last reasoning of the proof, so I will write all of it out.

Let's assume $p$ is not a prime i.e. $p = qs$. Since we assumed that $F_p$ is a field, than each element different from $0$ has a multiplicative inverse, i.e. $ql=1(mod p)$. From here, we can deduce that $p$ is divisor of $ql-1$. From here, we can deduce that $q$ also has to be divisor of $ql-1$.

I understand everything until the following point:
"The last statement is possible only if $q=1$." This means that $p$ must be a prime.

Why is it that if $q|ql-1$, then $q=1$?

Answer 1

A field, in particular, is a domain: i.e. has no zero divisors. If p is not prime $p= qs$ then $s,q \in F $ and both are not zero. Then $qs=p=0(modp)$ a contradiction since F is a domain.

Answer 2

Notice: $$q \ | \ ql-1 \Longrightarrow q = (q,ql-1) = (q,-1) = 1 \Longrightarrow q=1$$