show that $\operatorname{lcm}(n,n+1,\cdots, n+k) = rn{n+k\choose k}$ 
Let $n,k$ be positive integers. Show that $lcm(n,n+1,\cdots, n+k) = rn{n+k\choose k}$ for some positive integer r.

Observe that it suffices to show that $n{n+k\choose k}$ divides the lcm. For a prime p and positive integer m, let $e_p(m)$ denote the exponent of p in the prime factorization of m. One way to show that $n{n+k\choose k} = \dfrac{n(n+1)\cdots (n+k)}{k!}$ divides the lcm is to show that for all primes p, $e_p(n{n+k\choose k})$ is at most $e_p(lcm(n,n+1,\cdots, n+k)) = \max\{ e_p(n+i) :0\leq i\leq k\}.$ Let p be a prime and let $p^b$ be the highest exponent of p dividing any of $n+i$ for $0\leq i\leq k$ and suppose $n+u$ is divisible by $p^b$. Then $n+u + i \equiv i\mod p^b$ for $0\leq i\leq k-u.$ But does this imply that $i$ and $n+u+i$ have the same highest exponent of $p$? Also, how can I show that the exponent of p in $n(n+1)\cdots (n+k)$ does not exceed the exponent of p in $k!$ by more than $b$?

Note that once we prove the above claim, by taking $n=rk!$ for any positive integer r, we have $n{n+k\choose k} = rn(n+1)\cdots (n+k)$, which is a multiple of $n(n+1)\cdots (n+k)$. Since this common multiple divides the lcm, it must equal the lcm. Hence for infinitely many integer n, $lcm(n,n+1,\cdots, n+k)=n{n+k\choose k}.$


Edit: @Peter suggested using an induction over k. For $k=1,lcm(n,n+1,\cdots, n+k) = n(n+1)$ and $r=1$. Assume the result holds for k. Then ${n+k+1\choose k+1} = {n+k\choose k} + {n+k\choose k+1}.$ We need to consider for each prime p, what its maximum exponent is among $n,n+1,\cdots, n+k+1$ and compare this to the exponent of p in $n(n+1)\cdots (n+k)/k!$

 A: Fix $n,k$, where $n$ is a positive integer and $k$ is a nonnegative integer.

Let $L=\operatorname{lcm}(n,...,n+k)$.

Claim:$\;n{\large{{n+k\choose k}{\,\mid\,}L}}$.

Proof:

The claim holds if and only if the inequality
$$
v_p
\Bigl(
n{\small{{n+k\choose k}}}
\Bigr)
\le
v_p(L)
$$
holds for all primes $p$, where $v_p$ is the usual $p$-adic valuation.

With the goal of applying the above criterion, fix a prime $p$.

Letting $e=\max\bigl(v_p(n),...,v_p(n+k)\bigr)$, it follows that $v_p(L)=e$.

Let $r\in\{0,...,k\}$ be such that $v_p(n+r)=e$.

If $k+1\ge p^{e+1}$, then at least one of the $k+1$ consecutive integers $n,...,n+k$ would be divisible by $p^{e+1}$, contrary to the definition of $e$.

Hence $k+1 < p^{e+1}$, so also $k < p^{e+1}$.

Recall that for nonzero integers $a,b$, we have

*

*$v_p(a+b)\ge \min\bigl(v_p(a),v_p(b)\bigr)$.$\\[4pt]$

*$v_p(a+b)=\min\bigl(v_p(a),v_p(b)\bigr)$ if $v_p(a)\ne v_p(b)$.

Let $s$ be a nonzero integer such that $0\le r+s\le k$.

Then $|s|\le k$, hence $v_p(s)\le e$.

If $v_p(s)=e$ then
$$
e
\ge
v_p\bigl(n+(r+s)\bigr)
=
v_p\bigl((n+r)+s\bigr)
\ge
\min
\bigl(
v_p(n+r),v_p(s)
\bigr)
=
e
$$
hence $v_p\bigl((n+r)+s\bigr)=e=v_p(s)=v_p(|s|)$.

If instead we have $v_p(s) < e$, then $v_p(s) < v_p(n+r)$, hence
$$
v_p\bigl((n+r)+s\bigr)
=
\min
\bigl(
v_p(n+r),v_p(s)
\bigr)
=
v_p(s)
=
v_p(|s|)
$$
Thus in either case we have $v_p\bigl((n+r)+s\bigr)=v_p(|s|)$.

Recall that for nonzero integers $a,b$, we have

*

*$v_p(ab)=v_p(a)+v_p(b)$.$\\[4pt]$

*$v_p\Bigl({\large{\frac{a}{b}}}\Bigr)=v_p(a)-v_p(b)$ if $b{\,\mid\,}a$.

Then we get
\begin{align*}
&
v_p
\Bigl(
n{\small{{n+k\choose k}}}
\Bigr)
\le
v_p(L)
\\[4pt]
\iff\;&
v_p
\left(
\prod_{i=0}^k (n+i)
\right)
-
v_p(k!)\le e
\\[4pt]
\iff\;&
\sum_{i=0}^k
v_p(n+i)
\le
e+v_p(k!)
\\[4pt]
\iff\;&
\left(
\sum_{i=0}^{r-1}
v_p(n+i)
\right)
+
\;v_p(n+r)
+
\left(
\sum_{i=r+1}^k
v_p(n+i)
\right)
\le
e+v_p(k!)
\\[4pt]
\iff\;&
\left(
\sum_{i=0}^{r-1}
v_p(n+i)
\right)
+
\left(
\sum_{i=r+1}^k
v_p(n+i)
\right)
\le
v_p(k!)
\\[4pt]
\iff\;&
\left(
\sum_{s=1}^r
v_p\bigl((n+r)-s\bigr)
\right)
+
\left(
\sum_{s=1}^{k-r}
v_p\bigl((n+r)-s\bigr)
\right)
\le
v_p(k!)
\\[4pt]
\iff\;&
\left(
\sum_{s=1}^r
v_p(s)
\right)
+
\left(
\sum_{s=1}^{k-r}
v_p(s)
\right)
\le
v_p(k!)
\\[4pt]
\iff\;&
v_p\bigl(r!\bigr)
+
v_p\bigl((k-r)!\bigr)
\le
v_p(k!)
\\[4pt]
\iff\;&
v_p\Bigl(r!(k-r)!\Bigr)
\le
v_p(k!)
\\[4pt]
\iff\;&
v_p(k!)
-
v_p\Bigl(r!(k-r)!\Bigr)
\ge
0
\\[4pt]
\iff\;&
v_p{k\choose r}\ge 0
\\[4pt]
\end{align*}
which is true.

This completes the proof.
A: Let, for $j \in \{0,\dots,k\}$, $N_j^{(0)} := \displaystyle\prod\limits_{\substack{i = 0\\ i \neq j}}^{k}(n+i)$.
By using the same type of argument as in this post, we can show that:
$$\operatorname{lcm}(n,n+1,\dots,n+k) = \frac{\displaystyle\prod_{i=0}^{k}(n+i)}{\operatorname{gcd}\left(N_0^{(0)},\dots, N_k^{(0)}\right)}$$
To have the desired result, it suffices to check that there exists an integer $r$ such that $r\operatorname{gcd(\dots)} = k!$, since we'd then get:
$$rn \binom{n+k}{k} = r\,\cdot\,\frac{\displaystyle\prod_{i=0}^{k}(n+i)}{k!} = \frac{\displaystyle\prod_{i=0}^{k}(n+i)}{\operatorname{gcd}\left(N_0^{(0)},\dots, N_k^{(0)}\right)}$$
Let $d := \operatorname{gcd}\left(N_0^{(0)},\dots, N_k^{(0)}\right)$. Suppose by contradiction that $d$ does not divide $k!$.
Let, for $p$ prime, $v_p$ denote the usual $p$-adic valuation.
$d$ not dividing $k!$ implies that there exists $p$ prime such that: $$v_p(d) \geq v_p(k!)+1 = \left(\sum_{l = 1}^{k}v_p(l)\right) + 1$$
By definition of $k!$ and due to the fact that $v_p(ab) = v_p(a) + v_p(b)$.
Because $d$ is a common divisor of all the $N_{j}^{(0)}$, $p^{v_p(d)}$ does too, and we necessarily have that, for $j \in \{0,\dots,k-1\}$:
$$\begin{split}p^{v_p(d)} \mid N_j^{(0)} - N_{j+1}^{(0)} & = \displaystyle\prod\limits_{\substack{i = 0\\ i \neq j}}^{k}(n+i) \,\,- \displaystyle\prod\limits_{\substack{i = 0\\ i \neq j+1}}^{k}(n+i)\\
& = \left(\displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1\}}}^{k}(n+i)\right)\Big((n+j+1) - (n+j)\Big)\\
& = \left(\displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1\}}}^{k}(n+i)\right) \cdot 1 = \displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1\}}}^{k}(n+i) =: N_j^{(1)} \end{split}$$
Now, since $p^{v_p(d)} = p^{v_p(d) - v_p(1)}$ is a common divisor of all the $N_j^{(1)}$, we similarly get that, for $j \in \{0,\dots,k-2\}$:
$$\begin{split}p^{v_p(d) - v_p(1)} \mid N_j^{(1)} - N_{j+1}^{(1)} & = \displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1\}}}^{k}(n+i) \,\,- \displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j+1,j+2\}}}^{k}(n+i)\\
& = \left(\displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1,j+2\}}}^{k}(n+i)\right)\Big((n+j+2) - (n+j)\Big)\\
& = \left(\displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1,j+2\}}}^{k}(n+i)\right) \cdot 2 =: N_j^{(2)} \cdot 2\end{split}$$
Therefore, $p^{v_p(d) - v_p(1) - v_p(2)}$ divides all the $N_j^{(2)}$ once again.
We can then iterate the process repeatedly by defining, for $m \in \{0,\dots, k\}$ and $j \in \{0,\dots,k-m\}$, $N_j^{(m)} := \displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{j,j+1,\dots,j+m\}}}^{k}(n+i)$, and then looking at the differences $N_j^{(m)} - N_{j+1}^{(m)}$ for $j \in \{0,\dots,k-m-1\}$ and for $m \in \{0,\dots,k-1\}$.
We'll obtain that $p^{v_p(d) - \sum_{l=1}^{m} v_p(l)}$ divides $N_j^{(m+1)} \cdot (m+1)$, hence: $p^{v_p(d) - \sum_{l=1}^{m+1} v_p(l)} \mid N_j^{(m+1)}$ (and we are guaranteed to have integers at all times thanks to the starting assumption).
By induction, at the end of the step $m = k-1$, we'll have obtained that $p^{v_p(d) - \sum_{l = 1}^{k} v_p(l)}$ divides $N_0^{(k)} = \displaystyle\prod\limits_{\substack{i = 0\\ i \not\in \{0,1,\dots,k\}}}^{k}(n+i) = 1$ by convention.
But this would be absurd since that would mean that $p^{v_p(d) - \sum_{l=1}^{k} v_p(l)} = 1$ and as such that $v_p(d) - \sum_{l=1}^{k} v_p(l) = 0$ yet our assumption prevents that, thus $d$ divides $k!$, and we have finished by the previous observations.
A: You can find a proof in the paper "Nontrivial lower bounds for the least common multiple of some finite sequences of integers" by Bakir Farhi. It is Theorem $11$ on page $5$. Proofs are on section $3$. For your second claim on the equality $lcm(n,n+1,\ldots,n+k)=n{n+k \choose k}$, there it is proven that it suffices to be $n(n+k) \equiv 0 \pmod{k!}$.
I have found the proof using Approach Zero where a comment to one of the questions pointed to Wikipedia article on "Binomial coefficient" - section "Bounds and asymptotic formulas" where the above paper is cited.
