Regular expression for even/odd string on alphabet Given the alphabet $\{ a,b \}$, if I want to build the regular expression for even/odd lenght strings, I can think that an even lenght string is given by some combination of $a$'s and $b$'s, and the letters can be paired, so there are four possibilities: $aa,ab,ba,bb$, the regular expression(for even lenght strings) so is:
$(aa+ab+ba+bb)^* $
But can I write this in this other (more compact) way?
(Using, e.g., the notation of Hopcroft-Motwani-Ullman) given a language $L$,
we write $L^i:=\{ w_1w_2...w_i \ \ \ s.t. w_1,w_2,...,w_i \in L  \}$
Then given that $(a+b)$ is (as a regular expression) the notation for the strings formed by one symbol, $a$ or $b$, then $(a+b)^2$ as in the notation above is
$\{w_1w_2  \ \ \ s.t. w_1,w_2 \in (a+b)\}=\{aa,ab,ba,bb\}$
So i can rewrite the notation for the regular expression for even lenght words as:
$((a+b)^2)^*$
But this holds in general, so if I want the strings of lenght a multiple on $n \in \mathbb{N}$, I can consider:
$((a+b)^n)^*$
As the regular expression.
I feel like this is somehow cheating and that this argument is not valid, but to me it seems correct: because it is true that exponentiation is not an operation in the language of regular expressions, but it is also true that in reality here exponentiation is only the concatenation $n-$times of the regular expression $(a+b)$, so $(a+b)^2=(a+b)(a+b)$, $(a+b)^3=(a+b)(a+b)(a+b)$, which is always a valid operation for regular expressions.

So my question is: Is the first way in which I have expressed the
regular expressions for even-lenght words the most concise way to
write it? And if the second way in which I have expressed it is
correct, can this be generalised for $n$ generic (as I proposed)?
How about an expression for even strings $((a+b)\cdot(a+b))^*$ and deriving the one for odd strings from it: $((a+b)\cdot(a+b))^*\cdot(a+b)$?