Proof of $1+x\leq e^x$ for all x? Does anyone provide proof  of $1+x\leq e^x$ for all $x$?
What is the minimum $a(>0)$  such that $1+x\leq a^x$ for all x?

Well, The derivative of $1 + x$ is $1$ and the derivative of $e^x$ is $e^x$ so $e^x$ is increasing faster whenever $e^x > 1$.  Which happens whenever $x > 0$.  As $x = 0$, $1+x = 1$ and $e^x = 1$ so they are equal there but for all points of positive $x$, $e^x > 1 + x$ as it has increased more.
On the other hand for negative values of $x$, $e^x <1$ so $e^x$ would be increasing slower.  If $1+x$ where ever equal or greater then $e^x$, then $e^x$ would never have been able to "catch up".  So for negative $x$, $e^x > 1+x$ .
But at $x = 0$, $e^x = 1+x$.
