Show $\det(F_n)=1$ for all $n$ 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 $\det(F_1)=1$.
