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

Algebra foundations
Use mathematical induction to prove that $$\displaystyle{n}{\left({n}{2}+{5}\right)}$$ 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 $$\displaystyle{P}{\left({k}\right)}:{k}{\left({k}^{{{2}+{5}}}\right)}\text{ is }\div{i}\text{ is }\le\text{ by }{6}.{k}{\left({k}^{{{2}}}+{5}\right)}={6}{m}$$ for some m∈N. We have to prove that P(k+1) is also true.
$$P(k+1):(k+1)[(k+1)^{2+5}] =(k+1)[k^{2}+2k+1+5] =(k+1)[k^{2}+2k+6] =k^{3}+2k^{2}+6k+k^{2}*2k+6 =k^{3}+3k^{2}+8k+6 =k^{3}+5k+3k^{2}+3k+6 =k(k^{2}+5)+3(k^{2}+k+2) =(6m)+3(k^{2)+k+2)}$$
Since $$\displaystyle{k}^{{{2}}}+{k}+{2}$$ is divisible by 22 for every even and odd k. Therefore $$\displaystyle{3}{\left({k}^{{{2}}}+{k}+{2}\right)}$$ is divisible by 66 and hence $$\displaystyle{\left({6}{m}\right)}+{3}{\left({k}^{{{2}}}+{k}+{2}\right)}$$ is divisible by 6. Therefore, $$\displaystyle{P}{\left({k}+{1}\right)}$$ is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true.