Consider $f(x) = e^x x^n - (1-cx)$ where $n \ge 1$. $f(0) = -1$. $f(1) = e - 1 +c $, so if $c > 1-e$ there is a root between $0$ and $1$.
If $ c \le 1-e$, $f(x) = e^x x^n - 1 -|c|x$, so, since $e^x > 1+x$ for $x > 0$, if $x_0 = \max(1, |c|^{1/(n-1)})$, $x_0^{n-1}\ge |c|$ so $x_0^n \ge c x_0$ and $f(x_0) > 0$ so there is a root between $0$ and $x_0$.
In either case, we know an interval where there is a positive root.
If $c > 0$, then $f'(x) = e^x x^{n-1}(x+n) + c > 0$, so $f$ is increasing and there can be only one positive root and almost any root finder will do.
If $c \le 0$, we might have to do more, but it is late and I am tired.