1. If $\alpha_1, \ldots, \alpha_n$ are linearly independent algebraic numbers over $\rats$ , then $e^{\alpha_1}, \ldots, e^{\alpha_n}$ are algebraically independent over $\rats$ .
2. If $\alpha_1, \ldots, \alpha_n$ are distinct algebraic numbers over $\rats$ , then $e^{\alpha_1}, \ldots, e^{\alpha_n}$ are linearly independent over $\rats$ .

By induction hypothesis, suppose the statement is true when $n<k$ . Now suppose $n=k$ . If $\alpha_1, \ldots,\alpha_k$ are linearly independent over $\rats$ , $A_1,\ldots,A_k$ are algebraically independent and certainly linearly independent over $\rats$ . So suppose $\alpha_1,\ldots,\alpha_k$ are not linearly independent over $\rats$ . Without loss of generality, we can assume $s_1\alpha_1=s_2\alpha_2+\cdots+s_k\alpha_k$ , where $s_i\in\rats$ and $s_1 \neq0$ . By multiplying a common denominator we can further assume that $s_i\in\ints$ and $s_1>0$ . Then $$\pwr{A}{1}{s_1}=e^{s_1\alpha_1}=e^{s_2\alpha_2+\cdots+s_k\alpha_k}=e^{s_2\alpha_2}\cdots e^{s_k\alpha_k}= \pwr{A}{2}{s_2}\cdots\pwr{A}{k}{s_k}.$$ Since $r_1A_1=r_2A_2+\cdots+r_nA_n$ , we get \begin{eqnarray*} (r_1^{s_1})(\pwr{A}{2}{s_2}\cdots\pwr{A}{k}{s_k}) &=& (r_1^{s_1})\pwr{A}{1}{s_1} \\ &=& (r_1A_1)^{s_1} \\ &=& (r_2A_2+\cdots+r_kA_k)^{s_1} \\ &=& g(A_2,\ldots,A_k), \end{eqnarray*}where $g(x_2,\ldots,x_k)=(r_2x_2+\cdots+r_kx_k)^{s_1}\in\rats[x_2,\ldots,x_k].$ Partition the numbers $\pwr{s}{i}{'}s$ into non-negative and negative ones, so that, say, $s_{c(1)},\ldots,s_{c(l)}$ are non-negative and $s_{d(1)},\ldots,s_{d(m)}$ are negative, then $$(r_1^{s_1})\pwr{A}{c(1)}{s_{c(1)}}\cdots\pwr{A}{c(i)}{s_{c(i)}}=g(A_2,\ldots,A_k) \pwr{A}{d(1)}{s_{d(1)}}\cdots\pwr{A}{d(j)}{s_{d(j)}}.$$ If we define $$f(x_2,\ldots,x_k)=r_1^{s_1}\pwr{x}{c(1)}{s_{c(1)}}\cdots\pwr{x}{c(i)}{s_{c(i)}}-g(x_2,\ldots,x_k) \pwr{x}{d(1)}{s_{d(1)}}\cdots\pwr{x}{d(j)}{s_{d(j)}},$$ then $f(x_2,\ldots,x_k)\in\rats[x_2,\ldots,x_k]$ and $f(A_2,\ldots,A_k)=0.$ By the induction hypothesis, $f=0$ . It is not hard to see that $r_1=\cdots=r_k=0$ and therefore $A_1,\ldots,A_k$ are linearly independent.

$(1 \Longleftarrow 2).$ We first need two lemmas:

Lemma 1. Given 2., if $\alpha\neq 0$ is algebraic over $\rats$ , then $e^{\alpha}$ is transcendental over $\rats$ .

Proof. Suppose $f(e^\alpha)=0$ where $f(x)=r_0+r_1x+\cdots+r_nx^n\in\rats[x]$ . Then we have \begin{eqnarray*} 0 &=& r_0+r_1e^{\alpha}+\cdots+r_n(e^{\alpha})^n \\ &=& r_0e^0+r_1e^{\alpha}+\cdots+r_ne^{n\alpha}. \end{eqnarray*}Since $\alpha\neq0$ , $0,\alpha,\ldots,n\alpha$ are all distinct, $1, e^{\alpha},\ldots,e^{n\alpha}$ are linearly independent by the hypothesis. Thus, $r_0=r_1=\ldots=r_n=0$ and we have $f(x)=0$ , which means that $e^{\alpha}$ is transcendental over $\rats$ .

Lemma 2. Given 2., if $\alpha$ and $\beta$ are linearly independent and algebraic over $\rats$ , then $e^{\alpha}$ is transcendental over $\rats(e^{\beta})$ .

Proof. Let $A=e^{\alpha}$ and $B=e^{\beta}$ . Suppose $f(A)=0$ where $f(x)\in\rats(B)[x]$ . We want to show that $f(x)=0$ . Write $$f(x)=r_0(B)+r_1(B)x+\cdots+r_n(B)x^n,$$ where each $r_i(x)=p_i(x)/q_i(x)$ with $p_i(x)$ , $q_i(x)\neq 0 \in \rats[x]$ . Let $Q(x)=q_1(x)\cdots q_n(x)$ . So $Q(B)$ , being the product of the denominators $q_i(B)\neq 0$ , is non-zero. Multiply $f(x)$ by $Q(B)$ we get a new polynomial $g(x)$ such that $$g(x)=R_0(B)+R_1(B)x+\cdots+R_n(B)x^n,$$ where each $R_i(x)=r_i(x)Q(x)=p_i(x)Q(x)/q_i(x)\in \rats[x]$ . Now, $g(A)=f(A)Q(B)=0$ . So \begin{eqnarray*} 0 &=& R_0(B)+R_1(B)A+\cdots+r_n(B)A^n \\ &=& \sum_{j=0}^{m_0}a_{0j}B^j+\sum_{j=0}^{m_1}a_{1j}B^jA+\cdots+\sum_{j=0}^{m_n}a_{nj}B^jA^n \\ &=& \sum_{j=0}^{m_0}a_{0j}e^{j\beta}+\sum_{j=0}^{m_1}a_{1j}e^{j\beta+\alpha}+\cdots+ \sum_{j=0}^{m_n}a_{nj}e^{j\beta+n\alpha}, \end{eqnarray*}where each $a_{ij}\in \rats$ . Now, the exponents in the above equation are all distinct, or else we would end up with $\alpha$ and $\beta$ being linearly dependent, contrary to the assumption. Therefore, by 2 (Lindemann-Weierstrass Version 2), all $e^{i\beta+j\alpha}$ are linearly independent, which means each $a_{ij}=0$ . This implies that $g(x)=0$ . But $g(x)=f(x)Q(B)$ and $Q(B)\neq 0$ , we must have $f(x)=0$ .

Now onto the main problem. We proceed by induction on the number of linearly independent algebraic elements over $\rats$ . The case when $n=1$ is covered in Lemma 1, since a linearly independent singleton is necessarily non-zero. So suppose $\alpha_1, \ldots, \alpha_k$ are linearly independent and algebraic over $\rats$ . Then each pair $\alpha_k, \alpha_i$ are independent and algebraic over $\rats$ , $i\neq k$ . Thus $e^{\alpha_k}$ is transcendental over $\rats(e^{\alpha_i})$ for all $i\neq k$ . This means that $e^{\alpha_k}$ is transcendental over $\rats (e^{\alpha_1},\ldots,e^{\alpha_{k-1}}).$

Now let $A_i=e^{\alpha_i}$ for all $i=1,\ldots,k$ . Suppose $f(A_1,\ldots,A_k)=0$ where $f\in\rats[x_1,\ldots,x_k]$ . To show the algebraic independence of the $A_i's$ , we need to show that $f=0$ . Rearranging terms of $f$ and we have $$0 = f(A_1,\ldots,A_k) = \sum_{j=0}^{m}g_j(A_1,\ldots,A_{k-1})(A_k)^j.$$ If we let $g(x)=f(A_1,\ldots,A_{k-1},x)$ , we see that $g(x)\in\rats(A_1,\ldots,A_{k-1})[x]$ and $g(A_k)= f(A_1,\ldots,A_{k-1},A_k)=0$ . Since $e^{\alpha_k}$ is transcendental over $\rats(A_1,\ldots,A_{k-1})$ , we must have $g(x)=0$ . This implies that each $g_j(A_1,\ldots,A_{k-1})=0$ . But then $A_1,\ldots,A_{k-1}$ are algebraically independent by the induction hypothesis, we must have each $g_j=0$ . This means that $f=0$ .