The numbers $a_j$ do not have to be algebraic.

Theorem. If $a_j,\; 1\leq j\leq n$ are $Q$-linearly independent then $e^{a_jz}$
are algebraically independent over $C(z)$

Proof. Let $$F(x_1,\ldots,x_n)=\sum_j c_j(z)x_1^{m_{j,1}}\ldots x_n^{m_{j,n}},$$
where $c_j\in C(z)$, and $(m_{j,1}\ldots,m_{j,n})\neq(m_{k,1}\ldots,m_{k,n})$
for every pair $j\neq k$. Suppose that
$$F(e^{a_1z},\ldots,e^{a_nz})=\sum_jc_j(z)e^{z\sum_km_{j,k}a_k}\equiv 0.$$
All exponentials here are distinct, because $a_j$ are $Q$-linearly independent.
Then we have a contradiction from the asymptotics in the complex plane:
if there is only one exponent of the largest modulus, its growth dominates
the rest in certain directions. If there are several, their arguments are different and each dominates in certain direction.

Remark. Just noticed that this essentially coincides with ACL's comment:-)

Transcendental number theory, chapter 11 — The Siegel-Shidlovsky theorem. In particular, I quote the last sentence of §1: “Plainly also Theorem 11.1 includes Lindemann's theorem.” $\endgroup$ – ACL Nov 24 '16 at 14:05