TITLE: Is there a direct proof of this lcm identity?
QUESTION [36 upvotes]: The identity
$\displaystyle (n+1) \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \text{lcm}(1, 2, ... n+1)$
is probably not well-known.  The only way I know how to prove it is by using Kummer's theorem that the power of $p$ dividing ${a+b \choose a}$ is the number of carries needed to add $a$ and $b$ in base $p$.  Is there a more direct proof, e.g., by showing that each side divides the other?

REPLY [9 votes]: More generally, for $0 \leq k \leq n$, there is an identity
$(n+1) {\rm lcm} ({n \choose 0}, {n \choose 1}, \dots {n \choose k}) = {\rm lcm} (n+1,n,n-1, \dots n+1-k)$.  
This is simply the fact that any integer linear combination of $f(x), \Delta f(x), \Delta^2 f(x), \dots \Delta^k f(x)$ is an integer linear combination of $f(x), f(x-1), f(x-2), \dots f(x-k)$ where $\Delta$ is the difference operator, $f(x) = 1/x$, and $x = (n+1)$.