How to use other method to prove $2^x>2x-1$. let $x\in R$,show $$2^x>2x-1$$
My try:
 take a function $f(x)=2^x-2x+1$, then  we have
$$f'(x)=2^x\ln{2}-2,~~~~f''(x)=2^x\ln^2{2}>0$$
if let $x_{0}$ such   $f'(x_{0})=0\Longrightarrow 2^{x_{0}}=\dfrac{2}{\ln{2}}$ is  minimum the point (since $f''(x)>0$).
so we have
$$f(x)\ge f(x_{0})= \dfrac{2}{\ln{2}}-1+\dfrac{\ln{\ln{2}}}{\ln{2}}+1=\dfrac{2+\ln{\ln{2}}}{\ln{2}}>0$$
because $$2+\ln{\ln{2}}>0\Longleftrightarrow 2^{e^2}>e$$
have other  methods?Thanks 

$\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align} 
2^x&>2x-1
\tag{1}\label{1}
\end{align}
Let's consider
\begin{align} 
2^x&=2x-1
\tag{2}\label{2}
.
\end{align} 
It is known, that solution to equation \eqref{2}
can be found in terms of the
Lambert W function.
Let $2x-1=y$, $x=\tfrac12y+\tfrac12$:
\begin{align} 
2^{\tfrac12y+\tfrac12} &= y
\tag{3}\label{3}
,\\
\sqrt2\cdot 2^{\tfrac12y} &= y
,\\
\sqrt2\,\exp(\ln(2^{\tfrac12y})) &= y
,\\
\sqrt2\,\exp(\tfrac12y\cdot \ln2) &= y
,\\
y\,\exp(-\tfrac{\ln2}2\cdot y) &= \sqrt2
,\\
-\tfrac{\ln2}2\cdot y\,\exp(-\tfrac{\ln2}2\cdot y) &= -\tfrac{\ln2}2\cdot\sqrt2
,\\
\W\left(-\tfrac{\ln2}2\cdot y\,\exp(-\tfrac{\ln2}2\cdot y)\right) 
&=\W\left( -\tfrac{\ln2}2\cdot\sqrt2\right)
,\\
-\tfrac{\ln2}2\cdot y
&=\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{4}\label{4}
,\\
y
&=
-\tfrac2{\ln2}\cdot 
\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{5}\label{5}
,\\
x
&=
\tfrac12
-\tfrac1{\ln2}\cdot 
\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{6}\label{6}
.
\end{align} 
The argument of $\W\left( -\tfrac{\sqrt2}2\,\ln2\right)$
is $-\tfrac{\sqrt2}2\,\ln2<-\tfrac1{\mathrm e}$,
hence there are no real solutions to \eqref{2}, and the function
\begin{align} 
2^x-(2x-1)
\tag{7}\label{7}
\end{align} 
never cross the $x$-axis. Evaluation of \eqref{7} at $x=0$
gives 
\begin{align} 
2^0-(2\cdot0-1)
&=2> 0
\tag{8}\label{8}
,
\end{align} 
hence 
\begin{align} 
2^x-(2x-1)>&0
\quad\text{ for all } x\in\mathbb R
\tag{9}\label{9}
\end{align}
and \eqref{1} follows.
$\endgroup$
