# How to prove $n^5-5n^3+4n = 5!{n+2\choose 5}$

I understand the answer, but I am stuck on blue underline part. Please someone describe the blue underline section and how it's come from. Thank you.

Source

Number theory: Structure , Examples and Problems by Titu Andreescu .

• $$n(n^4\!-\!5n^2\!+\!4) = \color{#c00}n(\color{#0a0}{n^2\!-\!1})(n^2\!-\!4) = (n\!+\!2)(\color{#0a0}{n\!+\!1})\color{#c00}n(\color{#0a0}{n\!-\!1})(n\!-\!2) = \dfrac{(n\!+\!2)!}{(n\!-\!3)!} = 5!{n\!+\!2\choose 5}$$ Jul 13 at 16:52

$$5!\binom{n+2}{5} = 5!\frac{(n+2)!}{5!(n-3)!} = \frac{(n+2)!}{(n-3)!}$$$$= (n+2)(n+1)(n)(n-1)(n-2)=n^5-5n^3+4n.$$

Since $$\left( \begin{array}{l} n + 2 \\ \,\,\,\,5 \\ \end{array} \right) = \frac{{\left( {n + 2} \right)!}}{{\left( {n + 2 - 5} \right)!\,5!}}$$ it is $$\begin{array}{l} 5!\left( \begin{array}{l} n + 2 \\ \,\,\,\,5 \\ \end{array} \right) = \frac{{\left( {n + 2} \right)!}}{{\left( {n - 3} \right)!\,}} = \frac{{\left( {n + 2} \right)(n + 1)n(n - 1)(n - 2)(n - 3)!}}{{(n - 3)!}} = \\ \\ = \left( {n + 2} \right)(n + 1)n(n - 1)(n - 2) \\ \end{array}$$ that is what you need.