Expanding this, we get
\begin{align*}
[n]_5 &= n(n-1)(n-2)(n-3)(n-4) \\
&= n(n^4 - 10n^3 + 35n^2 - 50n + 24) \\
&= n^5 - 10n^4 + 35n^3 - 50n^2 + 24n.
\end{align*}
For $k=6$, we have $[n]_6 = n(n-1)(n-2)(n-3)(n-4)(n-5)$. Expanding this, we get
\begin{align*}
[n]_6
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