Bourbaki

Lemma 1

Tag 0A6Q, permanent. Cite it as [Bourbaki, Tag 0A6Q].

Lie Groups and Lie Algebras, chapter II, § 8. Convergence of the Hausdorff series (ultrametric case)

Let nn be an integer ⩾0\geqslant 0 and let n=n0+n1p+⋯+nkpkn = n_0 + n_1 p + \cdots + n_k p^k, with 0⩽ni⩽p−10 \leqslant n_i \leqslant p - 1, be the pp-adic expansion of nn. Let S(n)=n0+n1+⋯+nkS(n) = n_0 + n_1 + \cdots + n_k. Then

vp(n!)=n−S(n)p−1. v_p(n!) = \frac{n - S(n)}{p - 1}.
vp(n!)=∑i=1nvp(i)v_p(n!) = \sum_{i=1}^n v_p(i) and the number of integers ii between 1 and nn for which vp(i)⩾jv_p(i) \geqslant j is equal to the integral part [n/pj][n/p^j] of n/pjn/p^j. Then
vp(n!)=∑j⩾0j([n/pj]−[n/pj+1])=∑j⩾1[n/pj]. v_p(n!) = \sum_{j \geqslant 0} j ([n/p^j] - [n/p^{j+1}]) = \sum_{j \geqslant 1} [n/p^j].
As [n/pj]=∑i⩾jnipi−j[n/p^j] = \sum_{i \geqslant j} n_i p^{i-j}, the lemma follows.

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.