I am trying to solve the following exercise (page 133, from the book "Introduction to Vectors and Tensors, Vol. 1", where the link to the book can be found in http://rbowen.tamu.edu/). (Note that the equation numbers in parentheses are referring to the equation numbers in the book)
Problem: Show that the cofactor of an element of an $N\times N$ matrix $A$, define as
$$\mathrm{cof}A^s\,_t :=\varepsilon_{i_1 \dots i_N}A^{i_1}\,_1\dots A^{i_{t-1}}\,_{t-1}\delta^{i_t}\,_s A^{i_{t+1}}\,_{t+1} \dots A^{i_N}\,_{N}$$
can be written as
$$\mathrm{cof} A^{i_1}\,_{j_1}=\frac{1}{\left(N-1\right)!}\delta^{j_1\dots j_N}_{i_1\dots i_N}A^{i_2}\,_{j_2}\dots A^{i_N}\,_{j_N},$$
where delta denotes the generalized Kronecker delta, and implicit summation is assumed for every repeated index. This has to be derived from the definition of the cofactor in (21.18) or theorem (21.1), but I have trouble solving this.
Basically I want to use the following, easier problem and replicate the solution method for the above:
Problem: Show that the determinant of an $N\times N$ matrix, defined as
$$ \mathrm{det} A = \varepsilon_{i_1 \dots i_N}A^{i_1}\,_1 \dots A^{i_N}\,_N$$
can be written as
$$\mathrm{det}A=\frac{1}{N!}\delta^{j_1\dots j_N}_{1_1\dots i_N}A^{i_1}\,_{j_1}\dots A^{i_N}\,_{j_N}.$$
One can solve this (21.11) by using the following identity (21.8), which is derived from the definition of the determinant (21.1-3)
$$\varepsilon_{j_1\dots j_N}\mathrm{det}A=\varepsilon_{i_1\dots i_N}A^{i_1}\,_{j_1}\dots A^{i_N}\,_{j_N}$$
by multiplying (and summing) both side with $\varepsilon^{j_1\dots j_N}$ and using the identities $\varepsilon_{j_1\dots j_N}^{j_1\dots j_N}=N!$ and $\varepsilon^{j_1\dots j_N}\varepsilon_{i_1\dots i_N}=\delta^{j_1\dots j_N}_{i_1\dots i_N}$.
