The pattern looks pretty clear: you have
$\begin{align*} &\sum_{i=1}^ni=\frac12n(n+1)\\ &\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\ &\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;, \end{align*}\tag{1}$
where the righthand sides are closed formulas for the lefthand sides. Now you want
$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$
what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.
Added: The general result, of which the three in $(1)$ are special cases, is $\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$ For $n=1$ this is $k!=\frac1{k+1}(k+1)!\;,$ which is certainly true. Now suppose that $(2)$ holds. Then
$\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\ &\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\ &\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\ &\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\ &=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\ &=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;, \end{align*}$
exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.