Product of Stirling numbers of second kind

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What does the following expression evaluate to:

$$ k!{n\brace k} \cdot k!{n\brace k} $$

We know that $k!{n\brace k} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring takes us to a different power series or just to a different coefficient in the same power series?