Large Numbers Divisibility

Does $2^{2^{2011} + 1}$ divide $2^{2^{2012}} - 1$?

Is my solution correct?

Consider the ratio:

$$\frac{2^{2^{2012}} - 1}{2^{2^{2011}} + 1} = \frac{2^{2^{2011} \cdot 2^1} - 1}{2^{2^{2011}} + 1}$$

Let $2^{2011} = P$, then:

$$\frac{2^{P \cdot 2} - 1}{2^P + 1} = \frac{2^{2P} - 1}{2^P + 1}$$ $$\frac{\left(2^P\right)^2 - 1^2}{2^P + 1}$$ $$\frac{\left(2^P - 1\right)\left(2^P + 1\right)}{2^P + 1}$$ $$2^P - 1$$ $$2^{2^{2011}} - 1$$

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You could start by typing it up in $LaTeX$ so it is easier to read and doesn't take so much space. Please see meta.math.stackexchange.com/questions/107/… if you need help –  Ross Millikan Jan 20 '13 at 16:28
Yes it is correct. It is easy to verify: $\left(2^{2^{2011}} - 1\right) \left(2^{2^{2011}} + 1\right) = \left(2^{2^{2011}}\right)^2 - 1 = 2^{2^{2012}} - 1$ –  fran.aubry Jan 20 '13 at 17:12
@fran.aubry, Thank You Very Much for the help. A simple and beautiful way to verify. –  Ranjan Yajurvedi Jan 21 '13 at 1:10