$(12 \cdots n)$ in $S_n$ has $n-1$ inversions. What about the k-th composition?

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$\sigma = (12\cdots n)$ has $(n-1)$ inversions because it can be decomposed into transpositions $(1n)\cdots (13)(12)$ and $\sigma^k(i) = i + k \ (mod \ n)$. $\sigma^k$ has $k(n-k)$ inversions but how do we prove it?