Can anybody help me in understanding the underlined part in this image.
$q^k$ is coprime to $a^{n-q}$ becuase if it was not the case then $q|a^{n-q}$ which implies $q|a$ which is a contradiction to the fact that gcd$(a,n)=1$.
Is this argument correct?
We write $\binom{n}{q}=\frac{n!}{q!(n-q)!}=\frac{n(n-1)\ldots(n-q+1)}{q!}.$ The only number in the numerator that is divisible by $q$ is $n$ (it's a product of $q$ consecutive numbers). But $q^k|n$ and $q^{k+1}\not|n$, while $q|q!$ so $q^k$ cannot divide their quotient.
Your argument for $a^{n-q}$ is correct.