A simple proof is given in the <A HREF="https://artofproblemsolving.com/community/c7h1328834p7152628">Art of Problem Solving</A> (it is entered as an "olympiade problem").

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I reproduce the two-line proof for the record, with the change that $a_{11}\mapsto a_{nn}$:
$$\det{A}= \left| \begin{array}{cccccc}
a_{1,1}-a_{n,1}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,j}-a_{n,j}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,n-1}-a_{n,n-1}\dfrac{a_{1,n}}{a_{n,n}} & 0 \\ 
\dots & \dots & \dots & \dots & \dots & \dots \\ 
a_{i,1}-a_{n,1}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,j}-a_{n,j}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,n-1}-a_{n,n-1}\dfrac{a_{i,n}}{a_{n,n}} & 0 \\ 
\dots & \dots & \dots & \dots & \dots & \dots \\ 
a_{n-1,1}-a_{n,1}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,j}-a_{n,j}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,n-1}-a_{n,n-1}\dfrac{a_{n-1,n}}{a_{n,n}} & 0 \\ 
a_{n,1} & \dots & a_{n,j} & \dots & a_{n,n-1} & a_{n,n}
\end{array} \right|$$
$$=a_{n,n} \cdot \left| \begin{array}{ccccc}
\dfrac{b_{1,1}}{a_{n,n}} & \dots & \dfrac{b_{1,j}}{a_{n,n}} & \dots & \dfrac{b_{1,n-1}}{a_{n,n}} \\ 
\dots & \dots & \dots & \dots & \dots \\ 
\dfrac{b_{i,1}}{a_{n,n}} & \dots & \dfrac{b_{i,j}}{a_{n,n}} & \dots & \dfrac{b_{i,n-1}}{a_{n,n}} \\ 
\dots & \dots & \dots & \dots & \dots \\ 
\dfrac{b_{n-1,1}}{a_{n,n}} & \dots & \dfrac{b_{n-1,j}}{a_{n,n}} & \dots & \dfrac{b_{n-1,n-1}}{a_{n,n}}
\end{array} \right|= \dfrac{1}{{a_{n,n}}^{n-2}} \cdot \det{B} $$
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