In the textbook https://www.springer.com/gp/book/9783034851688 (Klassische elementare Analysis, by M. Koecher) the following elegant recurrence relation is proved for $\zeta(2n)$ (on p. 157): $$\left(n+\frac{1}{2}\right)\zeta(2n)=\sum\limits_{m=1}^{n-1}\zeta(2m)\,\zeta(2n-2m). \tag{1}$$
In fact (1) is equivalent to Euler's recurrence relation for Bernoulli numbers (independently found by Ramanujan) $$(2n+1)B_{2n}=-\sum\limits_{m=1}^{n-1}\binom{2n}{2m}B_{2m}\,B_{2n-2m}. \tag{2}$$ Why, In contrast to (2), (1) can seldom be found in the literature (I was able to find only https://link.springer.com/article/10.1007/s00591-007-0022-2 that mentions (1))? Are there any other references that discuss (1)?

In the textbook https://www.springer.com/gp/book/9783034851688 (Klassische elementare Analysis, by M. Koecher) the following elegant recurrence relation is proved for $\zeta(2n)$ (on p. 157): $$\left(n+\frac{1}{2}\right)\zeta(2n)=\sum\limits_{m=1}^{n-1}\zeta(2m)\,\zeta(2n-2m). \tag{1}$$
In fact (1) is equivalent to Euler's recurrence relation for Bernoulli numbers (independently found by Ramanujan) $$(2n+1)B_{2n}=-\sum\limits_{m=1}^{n-1}\binom{2n}{2m}B_{2m}\,B_{2n-2m}. \tag{2}$$ Why, In contrast to (2), (1) can seldom be found in the literature (I was able to find only https://link.springer.com/article/10.1007/s00591-007-0022-2 that mentions (1))? Are there any other references that discuss (1)?

P.S. In addition to juan's answer. G.T. Williams was not the first to state the result in this form. It can be found at least in N. Nielsen, Handbuch der theorie der gammafunktion, Leipzig: Druck und Verlag von B.G. Teubner, 1906, p. 49. I found this reference thanks to the paper "Some identities involving the Riemann zeta function. II." by R. Sitaramachandrarao and B. Davis, Indian J. Pure Appl. Math. 17(10):1175–1186, 1986. https://www.insa.nic.in/writereaddata/UpLoadedFiles/IJPAM/20005a50_1175.pdf This reference also has (1) and proves (among others) an interesting generalization of (1): $$4\sum\limits_{i+j+k=n}\zeta(2i)\zeta(2j)\zeta(2k)=(n+1)(2n+1)\zeta(2n)-6\zeta(2)\zeta(2n-2),$$ where $n\ge 3$ and the sum extends over all ordered triples $(i,j,k)$ of positive integers satisfying $i+j+k=n$.