Prove that $\det(A)=\det(A^T)$ algebraically

If you use the definition of determinant where you take the sum of all signed products of n elements where each element is chosen from a distinct row and distinct column, then you get that $\det(A) = \det(A^T)$ for free, and I guess you could call it an algebraic proof because you get the same algebraic formula for both $\det(A)$ and $\det(A^T)$.

For an $n\times n$ matrix $A=[A(i,j)]$ we define Determinant as$$\det A = \sum _{\sigma\in S_n} A(1,\sigma(1))\cdot A(2,\sigma(2))\cdots A(n,\sigma(n))$$

Above formula for $A$ would imply that $$\det A^T = \sum _{\sigma\in S_n} A^T(1,\sigma(1))\cdot A^T(2,\sigma(2))\cdots A^T(n,\sigma(n))$$ $$=\sum _{\sigma\in S_n} A(\sigma(1),1)\cdot A(\sigma(2),2)\cdots A(\sigma(n),n)$$

I guess you should do the remaining now which should not take much time to realize..

Caution : I am not saying that $$A(\sigma(1),1)\cdot A(\sigma(2),2)\cdots A(\sigma(n),n)=A(1,\sigma(1))\cdot A(2,\sigma(2))\cdots A(n,\sigma(n))$$ but their combined sum run over all permutations is same.

i.e., May be for some $\sigma_1, \sigma_ \in S_n$ you have $$A(\eta(1),1)\cdot A(\eta(2),2)\cdots A(\eta(n),n)=A(1,\sigma(1))\cdot A(2,\sigma(2))\cdots A(n,\sigma(n))$$