Proof of adjoint(ab) = adjoint(b)adjoint(a) So I'm trying to prove whether
$\operatorname*{adjoint}(AB) = \operatorname*{adjoint}(B)\operatorname*{adjoint}(A)$.
Here, for any matrix $C$, the matrix $\operatorname*{adjoint}(C)$ is defined as the complex conjugate of the transposed matrix of $C$.
My tactic is to look at the ijth entry of both matrices and compare, but the fact that AB is being transposed is confusing me. 
How should I do this proof? Is there a better way? 
 A: Let's look at the $(i,j)$th entry of $(AB)^\dagger$ (I use "dagger" for adjoint): 
$$
(AB)^\dagger_{i,j}=\overline{(AB)^T_{i,j}}=\overline{\sum_{k} A_{j,k}B_{k,i}}=\overline{\sum_{k}B_{k,i}A_{j,k}}=\sum_{k}\overline{B_{k,i}}\overline{A_{j,k}}=((B)^\dagger(A)^\dagger)_{i,j}
$$  To explain, the $(i,j)$th component of $(AB)^T$ will be the $(j,i)$th component of $AB$, which is the ("real") inner product of the $j$th row of $A$ with the $i$th column of $B$.  Putting the conjugate over top the sum, and simply re-ordering, we see that this is the same as taking the ("real") inner product of the $i$th column of $B$ with the $j$th row of $A$, in other words, taking the ("real") inner product of the $i$th row of $B^T$ with the $j$th column of $A^T$.  The conjugate also drops into the sum.
It's a lot of words but it comes down to this: rows become columns and the $(i,j)$th entry of a product is a (real) inner product of a row with a column.  I keep emphasizing "real" inner product to mean: 
$$
\sum_{i=1}^na_ib_i
$$ instead of the complex inner product which is: 
$$
\sum_{i=1}^na_i\overline{b_i}
$$
A: Note:I realised later, that OP means the conjugate transpose by transpose , Anyways, I've not deleted the work ,assuming the adjoint to be classical adjoint
I denote the conjugate transpose by superscript $H$,
For $A,B \in \mathcal{M}_n^n(\mathbb{C})$ we've $$(AB)^{H}=\overline{(AB)^{T}}=\overline{(B^{T}A^{T})}=\overline{B^{T}}\,\,\,\overline{A^{T}}=B^H\,\,A^{H}$$
.By definition for $A \in \mathcal{M}^n_n(\mathbb{R})$  we've $$A^{-1}=\frac{adj(A)}{det(A)}$$ Multiplying $A$ on both sides we get $$A^{-1}A=\frac{A\, adj(A)}{det(A)} \implies A\,adj(A)=I_{n}det(A)$$
Then we've $$AB\,adj(AB)= det(AB)I_{n}$$ Multipling $[adj(B)][adj(A)]$ on both sides we get $$adj(B)\,adj(A) \,AB\,adj(AB) =adj(B)\,adj(A)det(AB)I_{n}$$ $$ \implies adj(B)\,det(A)I_{n}\,B\,adj(AB)=det(AB)adj(B)\,adj(A)$$ $$ \implies det(A)adj(B)\,B\,adj(AB)=det(AB)adj(B)\,adj(A) $$ $$\implies det(A)det(B)I_{n}adj(AB)=det(AB)adj(B)\,adj(A) $$ $$\implies adj(AB)=adj(B)\,adj(A)$$
(Assuming $B \in \mathcal{M}^n_n(\mathbb{R})$)
