If K and Q are both groups and $h:Q\rightarrow \text{Aut}(K) $ is a homomorphism then the group operation for the semidirect product $K\rtimes_hQ$ is: $$(k_1,q_1)*(k_2,q_2)=(k_1h(q_1)(k_2),q_1q_2)$$
I am trying to prove that the inverse of $(k_1,q_1)\in K\rtimes_hQ $ is $(h(q_1^{-1})(k_1^{-1}),q_1^{-1})$.
I was able to prove it this way: $(k_1,q_1)*(h(q_1^{-1})(k_1^{-1}),q_1^{-1})=(1,1)$.
But I am having trouble proving it the opposite way:
$\begin{align}(h(q_1^{-1})(k_1^{-1}),q_1^{-1})*(k_1,q_1)&=(h(q_1^{-1})(k_1^{-1})h(q_1^{-1})(k_1),q_1^{-1}q_1)\\&=(h(q_1^{-1})(k_1^{-1})h(q_1^{-1})(k_1),1)\end{align}$
I'm not sure where to go from here. Any help would be appreciated, cheers.