Use mathematical induction to prove that n(n2+5) is divisible by 6 for n∈N

Let P(n) denotes that ${\displaystyle n\left(n^{n+5}\right)}$ is divisible by 6, for each natural number n.
${\displaystyle P\left(1\right):1\left(1^{2+5}\right)=6}$, which is divisible by 6. Therefore the result is true for n=1.
Now, assume that P(n) is true for n=k. That is ${\displaystyle P\left(k\right):k\left(k^{2+5}\right)\text{ is }÷i\text{ is }\le \text{ by }6.k(k^2+5)=6m}$ for some m∈N. We have to prove that P(k+1) is also true.
${\displaystyle P(k+1):(k+1)\left[{(k+1)}^{2+5}\right]}$

$=(k+1)[k^2+2k+1+5]$

$=(k+1)[k^2+2k+6]$

$=k^3+2k^2+6k+k^2\cdot 2k+6$

$=k^3+3k^2+8k+6$

$=k^3+5k+3k^2+3k+6$

$=k(k^2+5)+3(k^2+k+2)$

$=\left(6m\right)+3(k^2+k+2)$
Since ${\displaystyle k^2+k+2}$ is divisible by 22 for every even and odd k. Therefore ${\displaystyle 3(k^2+k+2)}$ is divisible by 66 and hence ${\displaystyle \left(6m\right)+3(k^2+k+2)}$ is divisible by 6. Therefore, ${\displaystyle P(k+1)}$ is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true.