Editing Lindemann–Weierstrass theorem (section)

====Final step====
We turn now to prove the theorem: Let ''a''(1), ..., ''a''(''n'') be non-zero [[algebraic number]]s, and ''α''(1), ..., ''α''(''n'') distinct algebraic numbers. Then let us assume that:

: <math>a(1)e^{\alpha(1)}+\cdots + a(n)e^{\alpha(n)} =  0.</math>

We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the ''a''(''i'')'s:

For every ''i'' ∈ {1, ..., ''n''}, ''a''(''i'') is algebraic, so it is a root of an [[irreducible polynomial]] with integer coefficients of degree ''d''(''i''). Let us denote the distinct roots of this polynomial ''a''(''i'')<sub>1</sub>, ..., ''a''(''i'')<sub>''d''(''i'')</sub>, with ''a''(''i'')<sub>1</sub> = ''a''(''i'').

Let S be the functions σ which choose one element from each of the sequences (1, ..., ''d''(1)), (1, ..., ''d''(2)), ..., (1, ..., ''d''(''n'')), so that for every 1&nbsp;≤&nbsp;''i''&nbsp;≤&nbsp;''n'', σ(''i'') is an integer between 1 and ''d''(''i''). We form the polynomial in the variables <math>x_{11},\dots,x_{1d(1)},\dots,x_{n1},\dots,x_{nd(n)},y_1,\dots,y_n</math>

: <math>Q(x_{11},\dots,x_{nd(n)},y_1,\dots,y_n)=\prod\nolimits_{\sigma\in S}\left(x_{1\sigma(1)}y_1+\dots+x_{n\sigma(n)}y_n\right).</math>

Since the product is over all the possible choice functions σ, ''Q'' is symmetric in <math>x_{i1},\dots,x_{id(i)}</math> for every ''i''. Therefore ''Q'' is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every ''i'', and in the variables ''y''<sub>''i''</sub>. Each of the latter symmetric polynomials is a rational number when evaluated in <math>a(i)_1,\dots,a(i)_{d(i)}</math>.

The evaluated polynomial <math>Q(a(1)_1,\dots,a(n)_{d(n)},e^{\alpha(1)},\dots,e^{\alpha(n)})</math> vanishes because one of the choices is just σ(''i'') = 1 for all ''i'', for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form

: <math>b(1)e^{\beta(1)}+ b(2)e^{\beta(2)}+ \cdots + b(N)e^{\beta(N)}= 0,</math>

where we already grouped the terms with the same exponent. So in the left-hand side we have distinct values β(1), ..., β(''N''), each of which is still algebraic (being a sum of algebraic numbers) and coefficients <math>b(1),\dots,b(N)\in\mathbb Q</math>.
The sum is nontrivial: if <math>\alpha(i)</math> is maximal in the lexicographic order, the coefficient of <math>e^{|S|\alpha(i)}</math> is just a product of ''a''(''i'')<sub>''j''</sub>'s (with possible repetitions), which is non-zero.

By multiplying the equation with an appropriate integer factor, we get an identical equation except that now ''b''(1), ..., ''b''(''N'') are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎

Note that Lemma A is sufficient to prove that ''e'' is [[irrational number|irrational]], since otherwise we may write ''e'' = ''p'' / ''q'', where both ''p'' and ''q'' are non-zero integers, but by Lemma A we would have ''qe''&nbsp;−&nbsp;''p'' ≠ 0, which is a contradiction. Lemma A also suffices to prove that {{pi}} is irrational, since otherwise we may write {{pi}} = ''k'' / ''n'', where both ''k'' and ''n'' are integers) and then ±''i''{{pi}} are the roots of ''n''<sup>2</sup>''x''<sup>2</sup> + ''k''<sup>2</sup> = 0; thus 2 − 1 − 1 = 2''e''<sup>0</sup> + ''e''<sup>''i''{{pi}}</sup> + ''e''<sup>−''i''{{pi}}</sup> ≠ 0; but this is false.

Similarly, Lemma B is sufficient to prove that ''e'' is transcendental, since Lemma B says that if ''a''<sub>0</sub>, ..., ''a''<sub>''n''</sub> are integers not all of which are zero, then

: <math>a_ne^n+\cdots+a_0e^0\ne 0.</math>

Lemma B also suffices to prove that {{pi}} is transcendental, since otherwise we would have 1&nbsp;+&nbsp;''e''<sup>''i''{{pi}}</sup>&nbsp;≠&nbsp;0.