Showing that the restricted mean survival time (RMST) of iid $\mathsf{exp}(\lambda)$ survival time random variables is decreasing in $\lambda$ I would like to show that $\frac{1-\exp(-\lambda\tau)}{\lambda}$ is a decreasing function of $\lambda$ where $\lambda,\tau>0$.
Motivation: Suppose we have iid continuous survival time random variables $T_i\sim\mathsf{exp}(\lambda)$. The restricted mean survival time (RMST), defined as the mean survival time of all subjects up to time $\tau$ is given by
$$\mathsf{RMST}_{\tau}(\lambda)=\int_0^{\tau} \exp(-\lambda t)dt=\frac{1-\exp(-\lambda\tau)}{\lambda}$$
We are interested in the posterior probability that $\mathsf{RMST}_{\tau}(\lambda)>\theta$ given our data. I would like to show that $\mathsf{RMST}_{\tau}(\lambda)>\theta$ for sufficiently small values of $\lambda$. Otherwise, if the function $\mathsf{RMST}_{\tau}(\lambda)$ were not monotone, it wouldn't make sense to simply evaluate the proportion of posterior samples $\hat{\lambda}$ such that $\mathsf{RMST}_{\tau}\left(\hat{\lambda}\right)>\theta$.
In order for the inequality $\mathsf{RMST}_{\tau}(\lambda)>\theta$ to hold for sufficiently small values of $\lambda$, the derivative
$$\frac{d}{d\lambda}\left(\frac{1-\exp(-\lambda\tau)}{\lambda}\right)=\frac{\exp(-\lambda\tau)\lambda\tau-1+\exp(-\lambda\tau)}{\lambda^2}$$
should be less than zero for all $\lambda>0$ and for some fixed $\tau>0$. However, it's not clear to me that this is the case. I believe it should be because $\exp(-\lambda t)$ is decreasing in $\lambda$ and we're integrating over positive values of $t$. How can I show this more rigorously? Are my concerns valid in that $\mathsf{RMST}_{\tau}(\lambda)$ should be monotone in order to make meaningful inference?

Since $\lambda^2 >0$ showing the derivative is negative reduces to showing that $\exp(-\lambda\tau)\lambda\tau-1+\exp(-\lambda\tau) = \exp (-\lambda \tau )(1 +\lambda \tau)  -1 \leq 0$
The right hand side is equivalent to $e^{-x}(1+x) \leq 1$  or $e^x \geq 1+x$
The function $g : x \mapsto e^x  -x - 1$ has derivative $e^x - 1 \geq 0$ for $x \geq 0$ hence $g$ is an increasing function with $g(0) = 0$ thus for $x \geq 0$, $g(x) \geq 0 \Rightarrow e^x \geq 1+x$
Then for $x \geq 0$, $e^{-x}(1+x) \leq 1$ and $\exp (-\lambda \tau )(1 +\lambda \tau)  -1 \leq 0$ if $\lambda \tau \geq 0$
