I know the answer is $n(n+1)$, but I'm having trouble formulating an argument. I know by the definition, if I let $h=[n,n+1]$ $$h=nk_1, h=(n+1)k_2$$ $$nk_1=(n+1)k_2$$ $$\frac{n}{n+1}=\frac{k_2}{k_1}$$ $$\frac{n(n+1)}{n+1}=\frac{(n+1)k_2}{k_1}$$ $$\frac{n(n+1)}{n+1}=\frac{h}{k_1}$$ $$n(n+1)=\frac{h(n+1)}{k_1}$$ So it appears the direct method is not working (unless I'm just not seeing it...)
I know if I start listing the multiples of each $$\{n, 2n, 3n, \ldots, (n-1)n, nn,(n+1)n\}$$ $$\{(n+1), 2(n+1), 3(n+1),\ldots, n(n+1), (n+1)(n+1)$$ My goal then would be to show there are no terms equal smaller than $n(n+1)$. So suppose suppose there is a smaller lcm. Then $$an=b(n+1), b<a$$ with $a<(n+1), b<n$. Then $an=bn+b<bn+a$
Thus, $an<bn+a \Rightarrow an-a<bn \Rightarrow a(n-1)<bn \Rightarrow \frac{a}{b}<\frac{n}{n-1}$ I am again stuck.
Probably the easiest way, as mm-aops suggests, is to use the general relationship $$ [m,n] = \frac{mn}{(m,n)}. $$ In this case, that reduces the problem to showing that $(n,n+1)=1$, which is easier.
But if you want to do it straight from the definition: the general multiple of $n+1$ is $k(n+1)$. What's the smallest $k$ for which is this a multiple of $n$? When $1\le k\le n-1$, should be able to convince yourself that the remainder when $k(n+1)$ is divided by $n$ is $k$. In particular, none of those are multiples of $n$.