This is an elaboration of Qiaochu Yuan's prior comment: there are complex solutions (in fact, infinitely many) to $e^t-1 = t$, and then $e^{tx}$ is a solution.
One root, the only real root, is $t=0$ which is actually a double root. Thus we have a two-term solution for this value of $t$, which is the familiar $y=ax+b$.
The other roots for $t$ are complex and so appear as conjugate pairs.
