1) ae a + be β + ... + se σ = 0 would be true if all coefficients a, b, , s were zero; it could be true with nonzero coefficients if any two exponents were equal; and it could be true, with nonzero coefficients and unequal exponents, if either the one or the other were transcendental. Thus 1 + e ιπ -0, and e (e) -e 2 = 0. Now, the Lindemann theorem says that the equation can be true in these cases only: LiNDEMANN THEOREM. The number e cannot satisfy an equation of the form (1), in which at

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... one coefficient a, b, ••• , s is different from zero, no two exponents (X, β, , σ are equal, and all numbers α, b 9 « , s; Gi, β, , σ are algebraic. The proof of this goes back to Hermite [2], Lindemann [5], and Weierstrass [10]. Since then the theorem has been a favorite topic for expository articles, alternative proofs, and so on, which turn up at the rate of several per decade. One naturally asks whether any good could come of still another note on this well-worn subject. What we have in mind here is the following. In [3] there is a discussion by Hubert of the transcendence of e and 77, in which the tools are chiefly analytic in character, and in which the proof for 77 is a simple extension of the proof for e. Hubert remarks that the general Lindemann theorem can be obtained in the same way. This line of inquiry was followed up by Klein [4" but in the authors' opinion the treatment is not complete. What appears to be the most delicate part is relegated to a footnote. Our primary objective, then, is to supply the details in the Hermite-Hilbert proof. Naturally we have drawn heavily on [3] and [4], and the whole treatment is expository in character. The exposition is made to depend on more elementary algebraic ideas than is the case, for example, in [7, pp. 1-24] and [8, p. 40] .