Let us denote $$k^{~\bar r}=k(k+1)(k+2)(k+3)...(k+r-1)$$ and $$k^{~\underline r}=k(k-1)(k-2)(k-3)...(k-r+1)$$
Then it is amusing to note that $$S_1=\sum_{k=1}^n k^{~\bar r}=\int_{0}^n k^{~\bar r} dk = \frac{n^{\overline{r+1}}}{r+1}=\frac{ n(n+1)(n+2)(n+3)...(n+r)}{r+1}.$$ $$S_2=\sum_{k=1}^n k^{~\underline r}=\int_{0 }^{n+1} k^{~\underline r} ~dk = \frac{(n+1)^{\underline{r+1}}}{r+1}=\frac{ (n+1)n(n-1)(n-2)(n-3)...(n-r-1)}{r+1}.$$ The question is: how to get $S_1$ and $S_2$ properly?
Your analogy can be made rigorous by introducing so-called indefinite and definite sums. The following is presented in more detail and nicely in section 2.6: Finite and infinite calculus in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
We start with the difference operator $\Delta$ defined as $\Delta f(x)=f(x+1)-f(x)$ and apply it on the falling factorial $x^{\underline{m}}$: \begin{align*} \color{blue}{\Delta\left(x^{\underline{m}}\right)}&=(x+1)^{\underline{m}}-x^{\underline{m}}\\ &=(x+1)x\cdots (x-m+2)\\ &\qquad-x(x-1)\cdots(x-m+1)\\ &=x(x-1)\cdots(x-m+2)\left[(x+1)-(x-m+1)\right]\\ &\color{blue}{=mx^{\underline{m-1}}}\tag{1} \end{align*} We observe in (1) a nice analogy between the delta operator $\Delta$ applied to the falling factorial $x^{\underline{m}}$ and the differential operator $D$ applied to the power function $x^m$. \begin{align*} Dx^m=mx^{m-1} \end{align*} We will use this analogy and start with introducing so-called definite sums: If $g(x)=\Delta f(x)$ we define the definite sum $\sum_{a}^b g(x)\delta x$ as \begin{align*} \color{blue}{\sum_{a}^b g(x)\delta x:= f(x)\Big|_a^{b}=f(b)-f(a)}\tag{2} \end{align*} Here we mimic definite integrals by recalling that if $g(x)=Df(x)$, then under appropriate conditions we have \begin{align*} \int_{a}^b g(x)\,dx=f(x)\Big|_{a}^{b}=f(b)-f(a) \end{align*} We see that $\delta$ in (2) plays the role of $d$ and $\Delta$ the role of $D$. Since \begin{align*} \color{blue}{\sum_{a\leq k<b}g(k)} &=\sum_{k=a}^{b-1}g(x)=\sum_{k=a}^{b-1}\Delta f(k)\\ &=\sum_{k=a}^{b-1}\left(f(k+1)-f(k)\right)\\ &\color{blue}{=f(b)-f(a)}\tag{3} \end{align*} we have from (2) and (3) \begin{align*} \color{blue}{\sum_{a}^{b}g(k)\delta k=\sum_{a\leq k<b}g(k)=\sum_{k=a}^{b-1}g(k)}\tag{4}\\ \end{align*}
Now its time to harvest. We obtain for positive integers $n$ and $r$: \begin{align*} \color{blue}{S_2}\color{blue}{=\sum_{k=1}^nk^{\underline{r}}} &=\frac{1}{r+1}\sum_{k=1}^n\Delta\left(k^{\underline{r+1}}\right)\tag{$\leftarrow$ (1)}\\ &=\frac{1}{r+1}k^{\underline{r+1}}\Big|_{1}^{n+1}\tag{$\leftarrow$ (2) - (4)}\\ &=\frac{1}{r+1}(n+1)^{\underline{r+1}}-\frac{1}{r+1}1^{\underline{r+1}}\\ &\color{blue}{=\frac{1}{r+1}(n+1)^{\underline{r+1}}} \end{align*} according to the claim.
In a similar way as above we get \begin{align*} \color{blue}{\Delta\left(x^{\overline{m}}\right)=m(x+1)^{\overline{m-1}}}\tag{5} \end{align*} and we calculate
\begin{align*} \color{blue}{S_1}\color{blue}{=\sum_{k=1}^nk^{\overline{r}}} &=\frac{1}{r+1}\sum_{k=1}^n\Delta\left((k-1)^{\overline{r+1}}\right)\tag{$\leftarrow$ (5)}\\ &=\frac{1}{r+1}\sum_{k=0}^{n-1}\Delta\left(k^{\overline{r+1}}\right)\\ &=\frac{1}{r+1}k^{\overline{r+1}}\Big|_{0}^{n}\tag{$\leftarrow$ (2) - (4)}\\ &\color{blue}{=\frac{1}{r+1}n^{\overline{r+1}}} \end{align*} according to the claim.
