$f(x)=e^x$ Prove $f[x_0,x_1,...,x_n]>0$

Question

The function is given that $$f(x)=e^x$$ Prove that the divided difference is positive for any distinct numbers $x_0,...x_n$ for $n\geq0$. First divided difference is $${ f[x_1]-f[x_0]\over x_1-x_0}$$ Now if $x_0>x_1$ $${ e^{x_1}-e^{x_0}\over x_1-x_0}>0$$ If $x_0<x_1$ $${ e^{x_1}-e^{x_0}\over x_1-x_0}>0$$ Hence, it is positive for any case for the first divided difference. Note that $x_0\neq x_1$ as all points are distinct. How can I prove that is it valid up to n th divided difference a.k.a $$f[x_0,x_1,...,x_n]>0$$ The divided difference here is defined as $$f[x_0,x_1,...,x_n]={f[x_1,...,x_{n}]-f[x_0,..x_{n-1}]\over x_n-x_0}$$