More reduction formulae troubles

[NewtonianAlch]

Homework Statement

\intx^{n}2^{x}dx = \frac{2}{ln 2} - \frac{n}{ln 2} \intx^{n-1}2^{x}dx

For n is greater or equal to 1, find \intx^{3}2^{x}dx

This is a definite integral from 0 to 1

The Attempt at a Solution

My first question is, after reducing this once, you are left with x^{2}, and attempting to reduce that again means the constant outside the integral will effectively become 0, therefore the constant term outside the integral will be 0 no matter how many more times you reduce it, and after you do the integral, 0 times the integrated result will be zero. What am I not seeing properly here?

Ignoring that and continuing to do the reduction, I get \frac{1}{ln 2}*\frac{1}{ln 2} which doesn't seem right to ignore it anyways.

That is to say, when it becomes x^{1}, and you apply the formula again, you'll get \frac{1}{ln 2} outside the integral, then you're left with x^{0} which is 1, and effectively you just integrate 2^{x} which becomes \frac{1}{ln 2}2^{x}, substituting 0 and 1, and doing the maths, you're left with \frac{1}{ln 2}*\frac{1}{ln 2} as the answer.

 

[Screwdriver]

I'm pretty sure that the formula is actually this (by using parts):

\int x^{n}2^{x}dx=\frac{2^{x} x^{n}}{ln(2)}-\frac{n}{ln(2)}\int x^{n-1}2^{x}dx

Applying this once:

\int x^{3}2^{x}dx=\frac{2^{x} x^{3}}{ln(2)}-\frac{3}{ln(2)}\int x^{3-1}2^{x}dx

http://www.wolframalpha.com/input/?i=integrate+x^%283%29+2^x+dx+%3D+%28%282^x+x^3%29%2Fln%282%29%29+-+%283+%2F+ln%282%29%29+integrate+x^%283+-+1%29+2^x+dx