Improving an upper bound of a function $f$ that satisfies an inequality involving $\exp(f)$.

Question

Let $f:(0,1] \to [0,M]$, $M>0$ be smooth and continuous in $[0,1]$ such that $f(t) \to 0$ as $t \to 0$. Furthermore, $f$ satisfies the following inequality $$ f(t) \leq C \cdot t \cdot \exp (f(t)) \quad \text{for all } t\in (0,1) $$ where $C>0$ does not depend on $t$. From this you can immediately get the weak estimate $$ f(t) \leq C \cdot t \cdot \exp (M) \quad \text{for all } t\in (0,1). $$ I believe that it should be possible to get a better estimate than this, at least for some suitable $t_0<1$. For since $f(t) \to 0$ as $t \to 0$ there is some $t_0$ such that $f(t)<\frac{1}{2}$ for all $t \in (0,t_0)$. Then the series expansion of the exponential function yields $$ \exp (f(t)) \leq 1+10 f(t) $$ for all $t \in (0,t_0)$. Then the first inequality becomes $$ f(t) (1 - 10 C t) \leq C t. $$ So for for $t \in (0, \min(t_0, \frac{1}{20C}))$ $$ f(t) \leq 2Ct. $$ In some way like this I need my inequality. However, my problem is, that now I don't have a clue what $t_0$ is. Is there a way to estimate a possible $t_0$ from below such that the last inequality holds? $t_0> \exp(-M)$ would be to bad. But $t_0> \frac{1}{C^mM^n}$ for some constant that does not depend on $M$ or $t_0$ would be good enough for me.