Question: I have a bit of trouble proving an adjugate matrix equality for $A\_{n\times n} = [a\_{ij}]$,

$$ \mbox{adj} (A) = \left[ \frac{\partial }{\partial a\_{ij} } \det(A) \right]^T, \quad i, j = 1,\dots,n $$

I tried by definition

$$ \mbox{adj} (A) = \left[
(-1)^{i+j} \sum\_{\sigma \in C\_n } \left( \mbox{sgn} (\sigma) \prod\_{k=1\\k\neq i }^n a\_{k,\sigma(k)} \right)
\right]^T
$$

where $C\_n = \{\sigma \in S\_n : \sigma(i) = j \} $ and $S\_n$ is the set of all $n$-permutation of the set $\{1,\dots,n\}$ . $sgn(\sigma)$ is $-1$ if permuatation $\sigma$ is odd , it's $1$ otherwise . $\sigma(k)$ returns the $k$th element of $\sigma$ . The thing within the square bracket is the $(i,j)-$cofactor . I pause here and look at the RHS of the statement

$$ \left[ \frac{\partial }{\partial a\_{ij} } \det(A) \right]^T 
= \left[\sum\_{\sigma\in C\_n } \left( \mbox{sgn} (\sigma) \prod\_{k=1\\ k \neq i}^n a\_{k,\sigma(k)} \right)
\right]^T $$

But I'm unable to deal with the signs ...

Answer: 
You can use Laplace expansion [wiki](https://en.wikipedia.org/wiki/Laplace_expansion) to write
$$
\det(\mathbf{A})
= \sum\_j C\_{ij} A\_{ij}
$$
where $C\_{ij}$ is called the cofactor of $A\_{ij}$.

It follows
$$
\frac{\partial \det(\mathbf{A})}{\partial A\_{ij}} = C\_{ij}
$$
or in matrix form
$$
\frac{\partial \det(\mathbf{A})}{\partial \mathbf{A}} 
= \mathbf{C}
= \left[\mathrm{adj}(\mathbf{A}) \right]^{T}
$$
Let's prove this identity in a slightly different way than the way you have tried to do. Let $\varphi^{\ell}$ be the only not decreasing bijection between the sets $\{1,2,\ldots, n-2,n-1\}$ and $\{1,2,\ldots,\ell-1,\ell+1,\ldots n-1,n\}$. Observe that for all fixed $\ell \in \{1,2,\ldots, n-1,n\}$ we have
$$
\varphi^{\ell}(1)\leq \varphi^{\ell}(2)\leq\ldots \leq \varphi^{\ell}(n-2)\leq\varphi^{\ell}(n-1)
$$
In this notation, the Laplace's expansion in $j$-column of $\det(A)$ is
$$
\det(A)=\sum\_{i=1}^{n} a\_{ij}\cdot (-1)^{i+j}\cdot \left( \sum\_{\sigma \in S^{n-1}}\prod\_{k = 1}^{n-1}a\_{\varphi^{i}(k)\sigma(\varphi^{j}(k))} \right)
$$
and
$$
\dfrac{\partial}{\partial \,a\_{ij}} \det(A) = (-1)^{i+j}\cdot \left( \sum\_{\sigma \in S^{n-1}}\prod\_{k = 1}^{n-1}a\_{\varphi^{i}(k)\sigma(\varphi^{j}(k))} \right)
$$
Now note that the determinant of the $(n-1)\times (n-1)$ submatrix of $A$ that is obtained by deleting row $i$ and column $j$ is given by
$
\left( \sum\_{\sigma \in S^{n-1}}\prod\_{k = 1}^{n-1}a\_{\varphi^{i}(k)\sigma(\varphi^{j}(k))} \right)
$. Then
$$
\mathrm{adj}(A)=\left( (-1)^{i+j}\cdot \left( \sum\_{\sigma \in S^{n-1}}\prod\_{k = 1}^{n-1}a\_{\varphi^{i}(k)\sigma(\varphi^{j}(k))} \right) \right)\_{n\times n}
$$