This is a problem from a students competition but it looks quite straightforward to solve.
The task is to compute the determinant of the following matrix.
A=[123...n122232...n2...............1n2n3n...nn]
What I noticed is that it looks similar to Vandermonde's matrix even though it's not quite that.
Let's pull out from its determinant a common multiple k from the k-th column for each k=1,2,3,...,n.
We get
det(A)=n!⋅|102030...n0112131...n1...............1n−12n−13n−1...nn−1|
The matrix whose determinant we got in the RHS of the last equality is the transpose of the Vandermonde matrix (with αs=s, for s=1,2,...,n). Hence its determinant equals the determinant of the Vandermonde matrix itself (since we know that det(A)=det(AT)).
So (leaving aside the n! multiplier) the determinant we are left with now is Vandermonde's determinant for the numbers 1,2,3,...,n. We know its value is equal to
∏1≤i<j≤n(j−i)
So the result we get now is
det(A)=n!⋅∏1≤i<j≤n(j−i)
This is the same as
n!⋅(n−1)1⋅(n−2)2⋅(n−3)3 ... 3n−3⋅2n−2⋅1n−1
Finally it is not difficult to realize that this expression is equal to
n! ⋅(n−1)! ⋅(n−2)! ... 3!⋅ 2!⋅ 1!
So this is our final answer here for the determinant of A.
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