Statement

Let α1<<αm\alpha_1<\cdots<\alpha_m be real. A nonzero f(x)=j=1mcjxαjf(x)=\sum_{j=1}^m c_jx^{\alpha_j} has at most m1m-1 distinct zeros in (0,)(0,\infty).

Inductive proof

The claim is immediate for one nonzero term. Divide by the smallest power actually present. The resulting function has a nonzero constant term; its derivative is a combination of at most m1m-1 distinct powers. If the original function had mm distinct positive zeros, Rolle's theorem would give at least m1m-1 zeros of this derivative, contradicting the inductive bound m2m-2. A derivative that vanished identically would leave a nonzero constant and hence no zeros.

Consequently the evaluation matrix [xjαi][x_j^{\alpha_i}] is invertible at distinct positive nodes. Otherwise a nonzero combination would vanish at all mm nodes. This is the generalized-power version of Vandermonde independence.