An identity with determinant and trace of a matrix

Question

How to prove the following identity:

$$\det(A)=\frac{1}{d!}\sum_{\sigma\in S_d}\mathrm{sgn}(\sigma)\mathrm{Tr}_{\sigma}(A)$$

where $\mathrm{Tr}_{\sigma}(A)$ is defined as following

if $\sigma$ is of type $1^{c_1}2^{c_2}\dots d^{c_d}$, then $\mathrm{Tr}_{\sigma}(A)=\prod_{i=1}^d(\mathrm{Tr}A^i)^{c_i}$

I see a proof using symmetric functions and Schur functors, but I'm not sure if I could explain it in less than 2 pages. On the other hand, I suspect that basic combinatorics should yield a simple proof. Observe that $\operatorname{Tr}\sigma\left(A\right) = \sum{i_1, i_2, \ldots, i_n} \prod_{j=1}^n a_{i_j, i_{\sigma\left(j\right)}}$. Here, $n$ is the size of the square matrix $A$ (I don't want to use your notation $d$ since I'm afraid you're using it in two different meanings). — darij grinberg, Jun 15 '15 at 11:27
Ah, yes! $\operatorname{Tr}\sigma\left(A\right) = \sum{i_1, i_2, \ldots, i_n} \prod_{j=1}^n a_{i_j, i_{\sigma\left(j\right)}} = \sum_{\tau : [n] \to [n]} \prod_{j=1}^n a_{\tau\left(j\right), \tau\left(\sigma\left(j\right)\right)}$, where $[n]$ is the set $\left{1,2,\ldots,n\right}$. Now, substitute this on the right hand side and put the summation over $\tau$ in front of the summation over $\sigma$. Then, notice that the inner sums where $\tau$ is not a permutation (i.e., not bijective) are $0$, while the remaining inner sums are $\operatorname{sgn}\left(\tau\right) \det A$. — darij grinberg, Jun 15 '15 at 11:28

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