If $a_1,\ldots,a_n$ are algebraic numbers, and $α_1,\ldots,α_n$ are distinct algebraic numbers, then$$a_{1}e^{\alpha _{1}}+a_{2}e^{\alpha _{2}}+\cdots +a_{n}e^{\alpha _{n}}=0$$has only the trivial solution $a_1 e^{\alpha_1} + a_2 e^{\alpha_2} + \cdots + a_n e^{\alpha_n} = 0$ has only the trivial solution $a_i = 0$ for all $i = 1, \dots, n.$

If $b(1),\ldots,b(n)$ are integers and $γ(1),\ldots,γ(n)$, are distinct algebraic numbers, then$$b(1)e^{\gamma(1)}+\cdots+ b(n)e^{\gamma(n)} = 0$$ has only the trivial solution $b(i)=0$ for all $i = 1, \dots, n.$

Let us choose a polynomial with integer coefficients which vanishes on all the $\gamma(k)$'s and let $\gamma(1),\ldots,\gamma(n),\gamma(n+1),\ldots,\gamma(N)$ be all its distinct roots. Let $b(n+1)=\ldots=b(N)=0$.

The polynomial$$P(x_1,\dots,x_N)=\prod_{\sigma\in S_N}(b(1) x_{\sigma(1)}+\cdots+b(N) x_{\sigma(N)})$$vanishes at $(e^{\gamma(1)},\dots,e^{\gamma(N)})$ by assumption. Since the product is symmetric, for any $\tau\in S_N$ the monomials $x_{\tau(1)}^{h_1}\cdots x_{\tau(N)}^{h_N}$ and $x_1^{h_1}\cdots x_N^{h_N}$ have the same coefficient in the expansion of $P$.

Thus, expanding $P(e^{\gamma(1)},\dots,e^{\gamma(N)})$ accordingly and grouping the terms with the same exponent, we see that the resulting exponents $h_1\gamma(1)+\dots+h_N\gamma(N)$ form a complete set of conjugates and, if two terms have conjugate exponents, they are multiplied by the same coefficient.

So we are in the situation of Lemma A. To reach a contradiction it suffices to see that at least one of the coefficients is non-zero. This is seen by equipping $C$ with the lexicographic order and by choosing for each factor in the product the term with non-zero coefficient which has maximum exponent according to this ordering: the product of these terms has non-zero coefficient in the expansion and does not get simplified by any other term. This proves Lemma B. ∎

For every $i∈\{1,\ldots,n\}$, $a(i)$ 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 $a(i)_1,\ldots,a(i)_{d(i)}$, with $a(i)_1=a(i)$.

Let $S$ be the functions $σ$ which choose one element from each of the sequences $(1,\ldots,d(1)),(1,\ldots,d(2)),\ldots,(1,\ldots,d(n))$, so that for every $1≤i≤n,σ(i)$ is an integer between 1 and $d(i)$. We form the polynomial in the variables $x_{11},\dots,x_{1d(1)},\dots,x_{n1},\dots,x_{nd(n)},y_1,\dots,y_n$ $$Q(x_{11},\dots,x_{nd(n)},y_1,\dots,y_n)=\prod\nolimits_{\sigma\in S}\left(x_{1\sigma(1)}y_1+\dots+x_{n\sigma(n)}y_n\right).$$ Since the product is over all the possible choice functions σ, $Q$ is symmetric in $x_{i1},\dots,x_{id(i)}$ 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 $y_i$. Each of the latter symmetric polynomials is a rational number when evaluated in $a(i)_1,\dots,a(i)_{d(i)}$. The evaluated polynomial $Q(a(1)_1,\dots,a(n)_{d(n)},e^{\alpha(1)},\dots,e^{\alpha(n)})$ 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 $$b(1)e^{\beta(1)}+ b(2)e^{\beta(2)}+ \cdots + b(N)e^{\beta(N)}= 0,$$ 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 $b(1),\dots,b(N)\in\mathbb Q$. The sum is nontrivial: if $\alpha(i)$ is maximal in the lexicographic order, the coefficient of $e^{|S|\alpha(i)}$ is just a product of $a(i)_j$'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 $b(1),\ldots,b(N)$ 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, since otherwise we may write $e=p/q$, where both $p$ and $q$ are non-zero integers, but by Lemma A we would have $qe−p$ ≠ 0, which is a contradiction. Lemma A also suffices to prove that $\pi$ is irrational, since otherwise we may write $\pi=kn$, where both $k$ and $n$ are integers) and then $±i\pi$ are the roots of $n^2x^2+k^2=0$; thus $2−1−1=2e^0+e^{i\pi}+e^{−i\pi}≠0$; but this is false.

Similarly, Lemma B is sufficient to prove that $e$ is transcendental, since Lemma B says that if $a_0,\ldots,a_n$ are integers not all of which are zero, then $$a_ne^n+\cdots+a_0e^0\ne 0.$$ Lemma B also suffices to prove that $\pi$ is transcendental, since otherwise we would have $1+e^{i\pi}≠0$.