F(x)=2^x; show (f(x+h)-f(x))/h=2^x{([2^h]-1)/â ¦

it's a basically a proof.i've gotten to the point where (2^x(x^h-1)/h) [f(x+h)-f(x))]/h
= [2^(x+h) - 2^x] / h
= 2^x (2^h -1)/h

note: 2^(x+h) = (2^x )(2^h)

Just so I know what I'm aiming for:
y = 2^x
ln y = x ln 2
y' / y = ln2
y ' = y ln2
y' = ln2 * 2^x

So by first principles:

f ' (x) = lim (h→0) (f(x + h) - f(x)) / h
= lim (h→0) (2^(x + h) - 2^x) / h
= lim (h→0) (2^x*2^h - 2^x) / h
= lim (h→0) 2^x(2^h - 1) / h

Your question just says to get to this bit....

= 2^x lim (h→0) (2^h - 1) / h

Now it can be shown that the lim bit = ln(2), but that's pretty tricky (see link below)