Bourbaki

Lemma 2

Tag 0A5E, permanent. Cite it as [Bourbaki, Tag 0A5E].

Lie Groups and Lie Algebras, chapter II, § 5. Magnus algebras

*Let nn be a non-zero rational integer. In the ring of formal power series K[[t]]K[[t]] we write (1+t)n=∑j≥0cj,ntj(1+t)^n = \sum_{j \geq 0} c_{j,n} t^j. There exists an integer j≥1j \geq 1 such that cj,n≠0c_{j,n} \neq 0.

If n>0n > 0, then cn,n=1c_{n,n} = 1 by the binomial formula. Suppose that n<0n < 0 and let m=−nm = -n. If cj,n=0c_{j,n} = 0 for all j≥1j \geq 1, then (1+t)n=1(1+t)^n = 1, whence, taking the inverse, (1+t)m=1(1+t)^m = 1, which contradicts the formula cm,m=1c_{m,m} = 1.*

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.