Simplification of powers I think this is a really simple question, but for some reason my brain can't get round it. I am proving a combinatorial result by probabilistic method and the last step has got me really confused.
Using indicator variables for an event with probability $1/2$ happening n times at once, I get probability $\left(\frac{1}{2}\right)^n=2^{-n}$. I also know that the amount of possible n element sets this occurs in is $<2^{1-n}.$ Thus the expectation of this event occurring becomes $<2^{-n}\cdot2^{1-n}$. I need to show that this is $<1$. 
This is where I am getting confused - the simplification of this!
I get:
$$2^{-n}\cdot2^{1-n}=2^{1-2n}=\frac{2^1}{2^{2n}}=\frac{2}{\left(2^2\right)^n}=\frac{2}{4^n}=\frac{1}{2^n}$$
But this is not correct. Can you explain why this doesn't work?
Also, I know that $n\geq1$ so I could say that $2^{1-2n}$ is always less than $1$ because $1-2n<1-2=-1$. Is this a valid statement instead? And would that be enough?
 A: $$\frac{2}{4^n} = \frac{2}{\underbrace{4\cdot 4\cdot 4\cdots 4}_{n~ \text{copies}}} = \frac{1}{2\cdot \underbrace{4\cdot 4\cdots 4}_{n-1~\text{copies}}} = \frac{1}{2\cdot 4^{n-1}}\neq \frac{1}{2^n}$$
Here, we simplify the fraction by cancelling the two on the top with part of one of the fours on the bottom.  You had somehow cancelled with all of the fours on the bottom which is invalid.
As for your question, yes $2^{1-2n}<1$ is true for all $n\geq 1$ since $2>1$ and $1-2n\leq -1$.  If all you needed to show was that much, then what you said is (mostly) fine (noting that $1-2n\leqq -1$ and not $1-2n< -1$ since the latter is untrue for $n=1$)
A: The short answer to your problem is that when you have something like $x^n$, it can be rewritten as $\frac{1}{x^{-n}}$. So 
$$2^{1-2n} = \frac{1}{2^{-(1-2n)}} = \frac{1}{2^{2n-1}}.$$
To work out the wrinkle in your logic, recall that raising something to an exponent works as a grouping symbol. When you have $\frac{2}{4^n}$, this is $2 / (4)^n$, and you can't reduce across the parenthesis because exponents come before division. It's kind of like how $(x + y)^2 \neq x^2 + y^2$ in general.
