I was reading through this answer to a question about the lower bound of a least common multiple.
The question is about showing that:
$$\text{lcm}(1,2,\dots,n) > 2^n$$
I was not clear on one step in the answer provided.
Here are the steps that I understand:
(1) I am clear on the definition of $I_{m,n}$ where $1 \le m \le n$:
$$I_{m,n}=\int_0^1x^{m-1}(1-x)^{n-m}dx = \sum_{r=0}^{n-m}\frac{a_r}{m+r}$$ where each $a_r \in \mathbb{Z}$
(2) I am clear that if $\displaystyle\ell_n = \text{lcm}(1,2,\dots,n)$ that:
$$\displaystyle\ell_n I_{m,n} \in \mathbb{Z}$$
I am unclear when Integration by Parts is applied.
(3) The answer states:
Now it can easily be seen that $I_{m,n} = \dfrac{1}{m{n\choose m}}$
Which follow from integration by parts or reduction formulae.
Could someone explain to me how this last step follows from the first two?
$I_{m,n}=\int_0^1 x^{m-1}(1-x)^{n-m+1-1}dx=B(m,n-m+1)=\cfrac{\Gamma(m) \Gamma(n-m+1)}{\Gamma(n+1)}$, which, if n and m are positive integers, is equal to $\cfrac{(m-1)!(n-m)!}{n!}=\cfrac{1}{m{n \choose m}}$, as claimed