Determinant with Unit Element in Otherwise Zero Row

Theorem
Let $$D$$ be the determinant:

$$D = \begin{vmatrix} 1 & 0 & \cdots & 0 \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \end{vmatrix}$$

Then $$D = \begin{vmatrix} b_{22} & \cdots & b_{2n} \\ \vdots & \ddots & \vdots \\ b_{n2} & \cdots & b_{nn} \end{vmatrix}$$.

Proof
We refer to the elements of $$\begin{vmatrix} 1 & 0 & \cdots & 0 \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \end{vmatrix}$$ as $$\begin{vmatrix}b_{ij}\end{vmatrix}$$.

Thus $$b_{11} = 1, b_{12} = 0, \ldots, b_{1n} = 0$$.

Then from the definition of determinant:

$$ $$

Now we note:

$$ $$

So only those permutations on $$\mathbb{N}^*_n$$ such that $$\lambda \left({1}\right) = 1$$ contribute towards the final sum.

Thus we have $$D = \sum_{\mu} \sgn \left({\mu}\right) b_{2 \mu \left({2}\right)} \cdots b_{n \mu \left({n}\right)}$$

where $$\mu$$ is the collection of all permutations on $$\mathbb{N}^*_n$$ which fix $$1$$.

Hence the result.