We have, $120=3 \times 5 \times 8$
Therefore to show that given expression is divisible by $120$ , it is sufficient to show that $3,5$ and $8$ are the factors of $n^{5}-5 n^{3}+4 n$
Let us factorise the given expression
$n^{5}-5 n^{3}+4 n=n\left[n^{4}-5 n^{2}+4\right]$
$=n\left[n^{4}-4 n^{2}-n^{2}+4\right]$
$=n\left[n^{2}\left(n^{2}-4\right)-1\left(n^{2}-4\right)\right]$
$=n\left(n^{2}-4\right)\left(n^{2}-1\right)$
$=n\left(n^{2}-2^{2}\right)\left(n^{2}-1^{2}\right)$
$=n(n-2)(n+2)(n-1)(n+1)$
$=(n-2)(n-1) n(n+1)(n+2)$
Since, $(n-2),(n-1), n,(n+1)$ are $4$ consecutive integers. One of it must be divisible by $4 .$ And at least two out of $5$ consecutive integers must be even. Thus, it must be divisible by $ 8 $.
Now, again out of three consecutive integers $1$ must be divisible by $3$, and out of $5$ consecutive integers l must be divisible by $5 .$
Thus, the given expression is divisible by $3,5,8$. Hence, it clear that $n^{5}-5 n^{3}+4 n$ is divisible by $120$ for all positive integer values of $n$.