$$
2^x > 1-x, \\
e^{x\ln2} > 1-x, \\
(1-x)e^{-x\ln2} < 1, \\
(1-x)e^{(1-x)\ln2} < e^{\ln 2} = 2^1, \\
(1-x)^2e^{\ln2} < e^{\ln 2}, \\
(1-x)^2 < 1, \\
x^2-2x+1-1 < 0, \\
x(x-2) < 0. \\
 $$
and then I did:
$$
x(x-2) = 0, \\
x\_1=0 , x\_2=2 \\
 $$
The solution is: $0<x<2 $ or $ x∈(0,2). \\$

I would like to know if my way of solving this inequallity is correct? (I do not have the solutions, so I do not know if $0<x<2 $ is the solution of this inequallity.)

Is there an easier way of solving this? Besides solving it graphically.

Thank you in advance!