How can I verify this family of values for hypergeometric functions?

Question

This Wolfram MathWorld page on hypergeometric functions states that

An infinite family of rational values for well-poised hypergeometric functions with rational arguments is given by $$_kF_{k-1}\left[\frac{1}{k+1},\dots,\frac{k}{k+1};\frac{2}{k},\frac{3}{k},\dots,\frac{k-1}{k},\frac{k+1}{k};\left(\frac{x(1-x^k)}{f_k}\right)^k\right]=\frac{1}{1-x^k}$$ for $k=2,3,\dots$, $0\leq x\leq(k+1)^{-1/k}$, and $$f_k\equiv\frac{k}{(1+k)^{1+1/k}}$$ (M. L. Glasser, pers. comm., Sept. 26, 2003).

So far, I've been unsuccessful in finding a source for this statement beyond "personal communication". Hypergeometric functions are only tangentially related to my problem, so I'm hoping someone more familiar with the topic might have seen this or a similar result. I am particularly interested in the case $x = (k + 1)^{-1/k}$.

Any help in tracking down the source of or verifying the statement is greatly appreciated.

Answer 1

The formula $$\sum_{n=0}^\infty \frac{1}{kn+1}\binom{(k+1)n}{n}(x(1-x)^k)^n=\frac{1}{1-x}\tag{1}$$ is well known and an easy consequence of Lagrange inversion. If $k$ is a positive integer $(1)$ may be written as $$_kF_{k-1}\left[\frac{1}{k+1},\dots,\frac{k}{k+1};\frac{2}{k},\frac{3}{k},\dots,\frac{k-1}{k},\frac{k+1}{k};\frac{(k+1)^{k+1}}{k^k} x(1-x)^k\right]=\frac{1}{1-x}.$$

Replacing $x$ with $x^k$ gives Glasser's formula.