ยง Geometric proof of e^x >= 1+x
, e^(x) >= 1x
Let's concentrate on the e^x >= 1 + x
part.
 The tangent of
e^x
at x = 0
is 1 + x
, since the taylor series of e^x
truncated upto x
is 1 + x
. 
e^x
is a strongly convex function, since (e^x)'' = e^x
which is positive everywhere. Hence, e^x
will always lie above its tangent.
Similarly for e^(x)
, working through the math:

1 x
is tangent at x=0
to e^(x)

(e^(x))'' = (e^(x))' e^(x)
which is again positive everywhere, and hence, e^(x)
is strongly convex.