Recently, I have been using the equation below to deduce many approximations to power series and the harmonic numbers, but I derived it only by recognizing the pattern for different values of $n$.
$$\sum_{k=m}^{\infty}\prod_{t=0}^{n}\frac{1}{(k+t)}=\frac{1}{n}\prod_{t=0}^{n-1}\frac{1}{(m+t)}$$
Is there a way to rigorously prove this identity? I thought about doing a partial fraction decomposition of the product, but I don't quite see how that would help.
Let $$f_n(k):=\prod_{t=0}^{n}\frac{1}{(k+t)}\,.$$ Then $$f_{n-1}(k)-f_{n-1}(k+1)=\frac{(k+n)-k}{\prod_{t=0}^{n} (k+t) }=nf_n(k)\,.$$ Thus we get a telescoping sum: $$\sum_{k=m}^{\infty} f_n(k)=\frac1n \sum_{k=m}^{\infty} \bigl(f_{n-1}(k)-f_{n-1}(k+1)\bigr)=\frac1n f_{n-1}(m) \,,$$ as required.