Power series identity 
Possible Duplicate:
Summation of $\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}$ 

Through a numerical computation, I stumbled across the following identity.  It takes place in the ring $(\mathbb{Z}[x])[[t]]$, which is complete with respect to the $t$-adic valuation.  The apparent identity is
$$\sum_{n=0}^\infty\frac{x(x+1)(x+2)\cdots(x+n-1)}{(1+t)(1+2t)\cdots(1+nt)}t^n=\frac1{1-xt}$$
I have numerically verified that this holds $\bmod t^{50}$.  Does anyone have any ideas about how to prove this identity?
 A: Your identity is a special case of Gauss's hypergeometric identity (see here for a proof),
$$\begin{align*}
\sum_{n=0}^\infty\frac{\prod_{j=0}^{n-1} (x+j)}{\prod_{j=0}^{n-1} (1+(j+1)t)}t^n&=\sum_{n=0}^\infty\frac{\prod_{j=0}^{n-1} (x+j)}{\prod_{j=0}^{n-1} \left(1+\frac1{t}+j\right)}\\
&=\sum_{n=0}^\infty \frac{(x)_n}{\left(1+\frac1{t}\right)_n}=\sum_{n=0}^\infty \frac{(x)_n (1)_n}{\left(1+\frac1{t}\right)_n}\frac1{n!}\\
&={}_2 F_1\left({{1,x}\atop{1+\frac1{t}}}\mid 1\right)=\frac{\Gamma\left(1+\frac1{t}\right)\Gamma\left(\frac1{t}-x\right)}{\Gamma\left(\frac1{t}\right)\Gamma \left(1+\frac1{t}-x\right)}
\end{align*}$$
where ${}_2 F_1\left({{a,b}\atop{c}}\mid z\right)$ is the Gaussian hypergeometric function, and since $\Gamma(1+z)=z\Gamma(z)$,
$$\require{cancel} {}_2 F_1\left({{1,x}\atop{1+\frac1{t}}}\mid 1\right)=\frac{\cancel{\Gamma\left(\frac1{t}\right)}\cancel{\Gamma\left(\frac1{t}-x\right)}}{t\left(\frac1{t}-x\right)\cancel{\Gamma\left(\frac1{t}\right)}\cancel{\Gamma \left(\frac1{t}-x\right)}}=\frac1{t\left(\frac1{t}-x\right)}=\frac1{1-xt}$$
