obtaining Bernoulli numbers from determinant I am reading a paper entitled Bernoulli Numbers Via Determinants by Hongwei Chen and I'm confused about a particular step.  The author sets up a system of equations via the following: first let $B_n$ represent the $n$-th Bernoulli number.  Then
$$x=(e^x-1)\sum_{n=0}^\infty B_n\frac{x^n}{n!}$$
Letting $b_n=B_n/n!$ and expanding we get
$$x=\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)\left(b_0+b_1x+b_2x^2+...\right)$$
Immediately we can see that $b_0=1$.  Using Cauchy Products, we then obtain an infinite sequence of equations which are coefficients of the powers of $x$. But we don't need infinite, we can look at the system for coefficients up to $x^n$.   Therefore
\begin{cases}
b_1=-\frac{1}{2!} \\[2ex]
\frac{b_1}{2!}+b_2=-\frac{1}{3!} \\[2ex]
\frac{b_1}{3!}+\frac{b_2}{2!}+b_3=-\frac{1}{4!} \\[2ex]
\vdots \\[2ex]
\frac{b_1}{n!}+\frac{b_2}{(n-1)!}+...+b_n=-\frac{1}{(n+1)!} \\[2ex]
\end{cases}
Then the author goes on to state that applying Cramers rule produces
$$b_n=
\begin{vmatrix}
1 & 0 & 0 & \cdots & -\frac{1}{2!} \\
\frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\
\frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\
\end{vmatrix}
$$
Then
$$B_n=n!
\begin{vmatrix}
1 & 0 & 0 & \cdots & -\frac{1}{2!} \\
\frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\
\frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\
\end{vmatrix}
$$
And finally, 
$$B_n=(-1)^n n!
\begin{vmatrix}
\frac{1}{2!} & 1 & 0 & \cdots & 0 \\
\frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & 0 \\
\frac{1}{4!} & \frac{1}{3!} & \frac{1}{2!} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{(n+1)!} &\frac{1}{n!} & \frac{1}{(n-1)!} &  \cdots & \frac{1}{2!} \\
\end{vmatrix}
$$
Can someone please explain the last step?  I just don't see how that last step comes from the previous.  I figure it has something to do with properties of determinants, and i do know that $|cA|=c^n|A|$, but I don't see where that is coming from here.

Here is the answer:
If $c$ is a constant and $A$ is an $n \times n$ matrix, then
$\det c A = c^n \det A$. Note that here $c$ multiplies all columns
not just the first column. Now, if $c$ multiplies only one column
then $\det [A_1, A_2, \cdots, c A_i, \cdots A_n]= c \det A$.
Now for the problem. The constant $c=-1$ only multiplies the first
column after being rolled from the last place ($n-1$ swaps).
Then we have the total sign is $(-1)^{n-1} (-1) = (-1)^n$.
