If $a\leq0$, then the problem is trivial. Assume that $a>0$. Suppose that there exist $k$ such that $x_k<0$, and let $k$ be the smallest such integer. $$x_k=1-a\cdot e^{x_{k-1}}<0\implies x_{k-1}>-\ln a$$$$x_{k-1}=1-a\cdot e^{x_{k-2}}>-\ln a\implies x_{k-2}<\ln\frac{1+\ln a}a$$ We know that $1+\ln a<a$ for all $0<a<1$, which implies that $x_{k-2}<\ln\frac{1+\ln a}a<0$. But $k-2<k$. This is a contradiction, because we defined $k$ to be the smallest integer such that $x_k<0$.

mofumofu wrote: Let $a\le 1$ be a real number. Sequence $\{x_n\}$ satisfies $x_0=0, x_{n+1}= 1-a\cdot e^{x_n}$, for all $n\ge 1$, where $e$ is the natural logarithm. Prove that for any natural $n$, $x_n\ge 0$. Let $a$ be the real number not greater than $1 .$ Sequence $\{x_n\}$ satisfies $x_0=0, x_{n+1}= 1-a\cdot e^{x_n}$, for all $n\ge 0$, where $e$ is the base of natural logarithm. Prove that: For any natural number $n$, $x_n\ge 0.$

If \(a \leq 0\), it is obvious. Assume that \(a > 0\). We get, \(\implies 1-ae^{x_n} \geq 0\) \(\implies \ln{\frac{1}{a}} \geq x_n\) By induction, Base Case: \(x_0 = 0\) is obvious Induction Case: Assume true for \(1-ae^{x_k} \geq 0\) \(1-ae^{x_{k+1}} = 1-ae^{1-ae^{x_{k}}} \geq 1-ae^{1-ae^{\ln{\frac{1}{a}}}} = 1-a \geq 0\) Thus by induction our statement is true for all \(n \geq 0\).