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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?

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$$\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1} = (-y)^{x-1} (1-y)^n$$ 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, $$\sum_{k=0}^n \dbinom{n}k (-1)^{k} \dfrac1{x+k} = \beta(x,n+1)$$

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$$\sum_{k=0}^n \dbinom{n}k p^k = (1+p)^n$$

$$\sum_{k=0}^n \dbinom{n}k (-1)^k = (1-1)^n=0$$

$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} =\sum_{k=0}^n \binom{n}{k}(-1)^k (1-\frac{k}{x+k} )= \sum_{k=0}^n \binom{n}{k}(-1)^k -\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{k}{x+k}= $$

$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = -\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{k}{x+k}= \sum_{k=1}^n \binom{n}{k}(-1)^{k+1} \frac{k}{x+k} \tag1$$

$$ \prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n} $$

$$ A_1=(x+1)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-1}=\frac{n!}{(n-1)!}=n $$

$$ A_2=(x+2)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-2}=\frac{n!}{(-1)(n-2)!}=-n(n-1) $$

$$ A_3=(x+3)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-3}=\frac{n!}{(-1)(-2)(n-3)!}=+\frac{n(n-1)(n-2)}{2!} $$

$$ A_k=(x+k)\prod_{m=1}^n \frac{m}{x+m} \bigg|_{x=-k}=\frac{n!}{(-1)(-2)(-3)..(-(k-1))(n-k)!}=(-1)^{k+1}\frac{n!k}{1.2.3..(k-1).k(n-k)!}=(-1)^{k+1}\frac{n!k}{k!(n-k)!}=(-1)^{k+1}k\dbinom{n}k $$

$$ \prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n}=\sum_{k=1}^n \frac{A_k}{x+k}=\sum_{k=1}^n \binom{n}{k}(-1)^{k+1} \frac{k}{x+k} $$

If we use equation 1

$$ \prod_{k=1}^n \frac{k}{x+k}= \sum_{k=0}^n \binom{n}{k}(-1)^{k} \frac{x}{x+k} $$

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