A simple proof is given in the Art of Problem Solving (it is entered as an "olympiade problem").
I reproduce the two-line proof for the record, with the change that $a_{11}\mapsto a_{nn}$: $$\det{A}= \left| \begin{array}{cccccc} a_{1,1}-a_{n,1}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,j}-a_{n,j}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,n-1}-a_{n,n-1}\dfrac{a_{1,n}}{a_{n,n}} & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{i,1}-a_{n,1}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,j}-a_{n,j}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,n-1}-a_{n,n-1}\dfrac{a_{i,n}}{a_{n,n}} & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{n-1,1}-a_{n,1}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,j}-a_{n,j}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,n-1}-a_{n,n-1}\dfrac{a_{n-1,n}}{a_{n,n}} & 0 \\ a_{n,1} & \dots & a_{n,j} & \dots & a_{n,n-1} & a_{n,n} \end{array} \right|$$ $$=a_{n,n} \cdot \left| \begin{array}{ccccc} \dfrac{b_{1,1}}{a_{n,n}} & \dots & \dfrac{b_{1,j}}{a_{n,n}} & \dots & \dfrac{b_{1,n-1}}{a_{n,n}} \\ \dots & \dots & \dots & \dots & \dots \\ \dfrac{b_{i,1}}{a_{n,n}} & \dots & \dfrac{b_{i,j}}{a_{n,n}} & \dots & \dfrac{b_{i,n-1}}{a_{n,n}} \\ \dots & \dots & \dots & \dots & \dots \\ \dfrac{b_{n-1,1}}{a_{n,n}} & \dots & \dfrac{b_{n-1,j}}{a_{n,n}} & \dots & \dfrac{b_{n-1,n-1}}{a_{n,n}} \end{array} \right|= \dfrac{1}{{a_{n,n}}^{n-2}} \cdot \det{B} $$Add the $n$-th row of $A$ to the $i$-th row, multiplied by $-a_{in}/a_{nn}$: $$\det{A}= \left| \begin{array}{cccccc} a_{1,1}-a_{n,1}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,j}-a_{n,j}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,n-1}-a_{n,n-1}\dfrac{a_{1,n}}{a_{n,n}} & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{i,1}-a_{n,1}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,j}-a_{n,j}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,n-1}-a_{n,n-1}\dfrac{a_{i,n}}{a_{n,n}} & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{n-1,1}-a_{n,1}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,j}-a_{n,j}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,n-1}-a_{n,n-1}\dfrac{a_{n-1,n}}{a_{n,n}} & 0 \\ a_{n,1} & \dots & a_{n,j} & \dots & a_{n,n-1} & a_{n,n} \end{array} \right|$$ $$=a_{n,n} \cdot \left| \begin{array}{ccccc} \dfrac{b_{1,1}}{a_{n,n}} & \dots & \dfrac{b_{1,j}}{a_{n,n}} & \dots & \dfrac{b_{1,n-1}}{a_{n,n}} \\ \dots & \dots & \dots & \dots & \dots \\ \dfrac{b_{i,1}}{a_{n,n}} & \dots & \dfrac{b_{i,j}}{a_{n,n}} & \dots & \dfrac{b_{i,n-1}}{a_{n,n}} \\ \dots & \dots & \dots & \dots & \dots \\ \dfrac{b_{n-1,1}}{a_{n,n}} & \dots & \dfrac{b_{n-1,j}}{a_{n,n}} & \dots & \dfrac{b_{n-1,n-1}}{a_{n,n}} \end{array} \right|= \dfrac{1}{{a_{n,n}}^{n-2}} \cdot \det{B} $$