<s>[INST] How do you write a polynomial as a linear combination of binomial coefficients? So here Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 in a proof, the needed polynomial is written as a linear combination of binomial coefficients, and I just can't figure out how. If some of you could give me a step by step explanation on how do you exactly turn a polynomial into a form like that, I would be thankful.
 [/INST] Note the powers $n^k$ can also be expressed with binomial coefficients $\binom{n}{k}$ by means of the so-called Stirling numbers of the second kind denoted with
$\begin{Bmatrix}n\\k\end{Bmatrix}$.In the referred Wiki-page the relationship is given using falling powers
\begin{align*}
(n)_k=n(n-1)(n-2)\cdots(n-k+1)
\end{align*}
as
\begin{align*}
n^k=\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}(n)_j\tag{1}
\end{align*}

Since binomial coefficients $\binom{n}{j}=\frac{(n)_j}{j!}$ we can write (1) as
  \begin{align*}\
n^k=\sum_{j=0}^k\begin{Bmatrix}k\\j\end{Bmatrix}\binom{n}{j}j!\tag{2}
\end{align*}
In order to represent $n^5-5n^3+4n$ with binomial coefficients we calculate according to (2)
\begin{align*}
n^5&=\sum_{j=0}^5\begin{Bmatrix}5\\j\end{Bmatrix}\binom{n}{j}j!
=1\binom{n}{1}1!+15\binom{n}{2}2!+25\binom{n}{3}3!+10\binom{n}{4}4!+\binom{n}{5}5!\\
n^3&=\sum_{j=0}^3\begin{Bmatrix}3\\j\end{Bmatrix}\binom{n}{j}j!
=1\binom{n}{1}1!+3\binom{n}{2}2!+1\binom{n}{3}3!\\
n^1&=\sum_{j=0}^1\begin{Bmatrix}1\\j\end{Bmatrix}\binom{n}{j}j!
=1\binom{n}{1}1!\\
\end{align*}
Putting all together gives
  \begin{align*}
n^5-5n^3+4n&=120\binom{n}{5}+240\binom{n}{4}+140\binom{n}{3}+30\binom{n}{2}+\binom{n}{1}\\
&\qquad-5\left(6\binom{n}{3}+6\binom{n}{2}+\binom{n}{1}\right)\\
&\qquad+4\binom{n}{1}\\
&=120\left(\binom{n}{5}+2\binom{n}{4}+\binom{n}{3}\right)\tag{3}
\end{align*}
Applying the induction step $n\rightarrow n+1$ to (3) it is obvious that the expression is a multiple of $120$.

</s>