Prove $\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}$ and more The current issue (vol. 120, no. 6)
of the American Mathematical Monthly
has a proof by probabilistic means
that
$$\sum_{k=0}^n  \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}
$$
for all $x > 0$
and all
$n \in \mathbb{N}$.
The article also mentions two other ways to prove this,
one using hypergeometric functions and
the Chu-Vandermonde formula,
and the other using the Rice integral formula
and complex contour integration.
This made me wonder if there were
more elementary ways to prove this result,
and my question is a challenge 
to find the most elementary proof.
Two ideas that have occurred to me
are (1) using Lagrange interpolation
and (2) induction.
I have not yet completed a proof,
so I am putting the problem out here.
The article states with hints of proofs
the following results:
$$\sum_{k=0}^n  \binom{n}{k}(-1)^k \big(\frac{x}{x+k}\big)^2 
= \big(\prod_{k=1}^n \frac{k}{x+k}\big)\big(1+\sum_{k=1}^n \frac{x}{x+k}\big)
$$
and,
for $m\in \mathbb{N}$,
$$\sum_{k=0}^n  \binom{n}{k}(-1)^k \big(\frac{x}{x+k}\big)^m 
= \big(\prod_{k=1}^n \frac{k}{x+k}\big)
\big(1+\sum_{j=1}^{m-1} \sum_{1\le k_1 \le k_2 \le ... \le k_j \le n}
\frac{x^j}{\prod_{i=1}^j (x+k_i)}\big)
$$
What (relatively) elementary proofs
of these are there?
 A: $$\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1} = (-y)^{x-1} (1-y)^n$$
Hence,
$$\int_0^1\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1}dy = \int_0^1(-y)^{x-1} (1-y)^n dy$$
$$\sum_{k=0}^n \dbinom{n}k (-1)^{x+k-1}\int_0^1y^{x+k-1}dy = \sum_{k=0}^n \dbinom{n}k (-1)^{x+k-1} \dfrac1{x+k}$$
$$\int_0^1(-y)^{x-1} (1-y)^n dy = (-1)^{x-1}  \beta(x,n+1)$$
Hence,
$$\sum_{k=0}^n \dbinom{n}k (-1)^{k} \dfrac1{x+k} = \beta(x,n+1)$$
A: $$\sum_{k=0}^n \dbinom{n}k p^k = (1+p)^n$$
$$\sum_{k=0}^n \dbinom{n}k (-1)^k = (1-1)^n=0$$
$$\sum_{k=0}^n  \binom{n}{k}(-1)^k \frac{x}{x+k} =\sum_{k=0}^n  \binom{n}{k}(-1)^k (1-\frac{k}{x+k} )= \sum_{k=0}^n  \binom{n}{k}(-1)^k -\sum_{k=0}^n  \binom{n}{k}(-1)^k \frac{k}{x+k}= 
$$
$$\sum_{k=0}^n  \binom{n}{k}(-1)^k \frac{x}{x+k} = -\sum_{k=0}^n  \binom{n}{k}(-1)^k \frac{k}{x+k}= \sum_{k=1}^n  \binom{n}{k}(-1)^{k+1} \frac{k}{x+k}
\tag1$$
$$
\prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n}
$$
$$ 
A_1=(x+1)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-1}=\frac{n!}{(n-1)!}=n
$$
$$ 
A_2=(x+2)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-2}=\frac{n!}{(-1)(n-2)!}=-n(n-1)
$$
$$ 
A_3=(x+3)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-3}=\frac{n!}{(-1)(-2)(n-3)!}=+\frac{n(n-1)(n-2)}{2!}
$$
$$ 
A_k=(x+k)\prod_{m=1}^n \frac{m}{x+m} \bigg|_{x=-k}=\frac{n!}{(-1)(-2)(-3)..(-(k-1))(n-k)!}=(-1)^{k+1}\frac{n!k}{1.2.3..(k-1).k(n-k)!}=(-1)^{k+1}\frac{n!k}{k!(n-k)!}=(-1)^{k+1}k\dbinom{n}k
$$
$$
\prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n}=\sum_{k=1}^n   \frac{A_k}{x+k}=\sum_{k=1}^n  \binom{n}{k}(-1)^{k+1} \frac{k}{x+k}
$$
If we use equation 1
$$
\prod_{k=1}^n \frac{k}{x+k}= \sum_{k=0}^n  \binom{n}{k}(-1)^{k} \frac{x}{x+k}
$$
