Answer
$$\frac{{{d^n}}}{{d{x^n}}}\left( {{2^x}} \right) = {2^x}{\left( {\ln 2} \right)^n}$$
Work Step by Step
$$\eqalign{
& \frac{{{d^n}}}{{d{x^n}}}\left( {{2^x}} \right) \cr
& {\text{Calculating until the third derivative}} \cr
& \frac{d}{{dx}}\left( {{2^x}} \right) = {2^x}\ln 2 \cr
& \frac{{{d^2}}}{{d{x^2}}}\left( {{2^x}\ln 2} \right) = \ln 2\left( {{2^x}\ln 2} \right) = {2^x}{\left( {\ln 2} \right)^2} \cr
& \frac{{{d^3}}}{{d{x^3}}}\left( {{2^x}{{\left( {\ln 2} \right)}^2}} \right) = {\left( {\ln 2} \right)^2}\left( {{2^x}\ln 2} \right) = {2^x}{\left( {\ln 2} \right)^3} \cr
& {\text{Therefore, we can conclude that}} \cr
& \frac{{{d^n}}}{{d{x^n}}}\left( {{2^x}} \right) = {2^x}{\left( {\ln 2} \right)^n} \cr} $$
