Pumping Lemma proofs I am having a hard time writing a Pumping Lemma contradictory proof for the below statements. 
1) $L_1 = \{ ww \mid w \in E^* \}$ <--- I don't understand how to read this. This is what I tried: For any string (of any length) that can be cut down the middle there exists and element of any length? so I would have to pick a string of length X that when I split the string in half, it still can be any length? this is really confusing to try to prove, i'd appreciate any suggestions with this one.
2) $L_2 = \{ a^{2^n} \mid n \geq 0 \}$ <---- same with this, I don't understand what kind of string I would have to pick to disprove this.
Note: That is a Pumping Lemma is for Regular Languages.
I'd appreciate any hints or comments that may help me disprove these.
Many thanks in advance.
 A: For (1) the language is repeated strings from some alphabet. If $E$ was $\{0,1\}$ then $E^*$ is all binary strings and the language is $\{,00,11,0000,0101,1010,1111,000000,001001,\ldots\}$. It suffices to prove this language is not regular for $E=\{0,1\}$ since intersections of regular languages are regular. Now suppose the machine that recognizes it has $n$ states and consider the double of a binary string of length $n$: you must be able to pump one half of it and not the other.. so pick the string of length $n$ that looks like 1000[n zeros]001, but doing that breaks the property that it it is a double.
For (2) the language is $\{a,aa,aaaa,aaaaaaaa,aaaaaaaaaaaaaaaa,\ldots\}$.
Suppose it was recognized by a machine of $n$ states, consider the smallest $i$ such that $n \le 2^i$. In matching the string $a^{2^i}$ some substring must pump. Suppose that substring had length $k \le 2^i$. Then the machine must also match $a^{2^{i} + k}$ and $a^{2^{i} + k + k}$ one of which is not a power of two.
A: MHZ, maybe the url, http://www.comp.nus.edu.sg/~sanjay/cs4232/t4ans.pdf, will help you with some idea. See question 1, through (a) to (d). After reading it, maybe you konw how to prove it. :-)
