Consider the $n\times n$ matrix $F_n= (f_{i,j})$ of binomial coefficients $$f_{i,j}=\begin{pmatrix}i-1+j-1\\i-1\end{pmatrix}$$ Prove that $\det(F_n)=1$ for all $n$.
My current idea is to apply Leibniz formula for determinants and induction, but it seems too complicated. Any better ideas and suggestions are welcome.
Using the formula $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ you get, by applying the column operations $$ \left\{ \begin{array}{lcl} C_n&\gets& C_n-C_{n-1} \\[1mm] C_{n-1}&\gets& C_{n-1}-C_{n-2} \\[1mm] &\vdots\\[1mm] C_2&\gets& C_2-C_{1} \end{array} \right. $$ that $$\det(F_n) = \det\left( \begin{array}{c|ccc} 1&0&\cdots&0\\\hline *\\ \vdots&&F_{n-1}\\ *\end{array}\right) = \det(F_{n-1})$$ So that the sequence $\det(F_n)$ is constant equalt to its first term $\det(F_1)=1$.