So I have found a solution but I don't think it's the best one. So by saying that $$ |\sin x|\le 1 (\forall x \in R) \Rightarrow -1 \le \sin x \le 1 \Rightarrow 0 \le 1- \sin x$$ and if we add $$ +e^x $$ on both sides we get $$ e^x \le e^x +1 - \sin x $$ but because $$ e^x > 0 , \forall x \in R $$ we finally get: $$ 0 < e^x \leq e^x + 1 -\sin x \Rightarrow e^x+1-\sin x > 0$$
Is there a better way to prove this? Thanks for your time.
You can find the derivative, thats the other way i can suggest. Of the function $F(x)=e^x +1- \sin(x) $