Here is lower bound on $1-t$ which holds for all $0 < t < 1$ (and not just for $0 < t < 0.6$): \begin{align*} \ln(1-t) &= -t - \tfrac{1}{2}t^2 - \tfrac{1}{3}t^3 - \tfrac{1}{4}t^4 - \cdots\\ &> -t - \tfrac{1}{2} t^2 - \tfrac{1}{2} t^3 - \tfrac{1}{2} t^4 -\cdots\\ &= -t - \tfrac{1}{2}t^2 (1 + t + t^2 + \cdots)\\ &= -t - \tfrac{t^2}{2(1-t)}\ \ \ \ \mbox{ for }\ \ \ 0 < t < 1\,.. \end{align*} Exponentiating both sides gives that for all $0< t < 1$, \begin{equation} \label{eq:igal1} 1-t > \exp\left(-t-\tfrac{t^2}{2(1-t)}\right)\,. \end{equation} Consequently for every $0 < \epsilon < 1$ ($\epsilon$ can be arbitrarily close), we get that for all $0 < t < 1-\epsilon$, \begin{equation} \label{eq:igal2} 1-t > \exp\left( - t - \tfrac{1}{2 \epsilon} t^2\right). \end{equation} In particular, if $\epsilon = 0.4$, then it follows from \eqref{eq:igal2} that for all $0 < t < 0.6$ \begin{equation} \label{eq:igal3} 1-t > \exp\left( - t - 1.25 t^2\right) \end{equation} The bound \eqref{eq:igal1} is more tight than \eqref{eq:igal2}, and \eqref{eq:igal2} is more general than \eqref{eq:igal3}.