Proving the binomial coefficient identity $\binom{~s + t~ }{s} = \prod_{i=1}^s \prod_{j=1}^t \frac{i + j}{i + j - 1}$

I tried expanding the factorial, but I do not know how to finish the proof.

\begin{eqnarray*} \binom{~s + t~ }{s} & = & \frac{(s+t)!}{s! ~ t!}\\ & = & \frac{(s+t)(s+t-1) \cdots (t+2)(t+1)}{s!} \\ & = & \prod_{i=1}^s \frac{t + i}{i} \\ \end{eqnarray*}

How do I get $$\prod_{i=1}^s \frac{t + i}{i} = \prod_{i=1}^s \prod_{j=1}^t \frac{i + j}{i + j - 1}$$?

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–  Grigory M Oct 28 '12 at 10:17

Hint: Telescopic product. Write a few terms out for $j=1,2,3..$: $$\frac{i+1}i\cdot\frac{i+2}{i+1}\cdots\frac{i+t}{i+t-1} = \frac{i+t}i$$

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Merci, Berci! I thought something like this would be the case, but had trouble writing it down. –  Jack Rousseau Oct 28 '12 at 9:45

$$\binom{~s + t~ - 1}{s - 1} + \binom{~s + t~ - 1}{s}$$

$$\prod_{i=1}^{s-1} \prod_{j=1}^{t} \frac{i + j}{i + j - 1} + \prod_{i=1}^s \prod_{j=1}^{t-1} \frac{i + j}{i + j - 1}$$

$$(\prod_{j=1}^{t} \frac{s + j - 1}{s + j} + \prod_{i=1}^s \frac{i + t - 1}{i + t}) \prod_{j=1}^{t} \prod_{i=1}^s \frac{i + j}{i + j - 1}$$

$$(\prod_{j=1}^{t} \frac{s + j - 1}{s + j} + \prod_{i=1}^s \frac{i + t - 1}{i + t}) \binom{s+t}{s}$$

$$(\frac{s}{s + t} + \frac{t}{s + t}) \binom{s+t}{s}$$

$$\binom{s+t}{s}$$

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–  Vÿska Oct 28 '12 at 21:53