Let $\Omega\subset \mathbb{R}^n$ is a bounded smooth domain and $1<p,q<\infty$ such that $1/p + 1/q =1$. I have understood that there is a theorem, which says that for every $F\in \big(L^p(\Omega)\big)^*$ there exists $v\in L^q(\Omega)$ such that $$F(u) = <u,F> = \int_{\Omega} u(x)v(x) \ d\mu(x) \text{ for all } u \in L^p(\Omega).$$ Hence we may associate $F$ to $v$.
Question: Is it common to write $<u,F> = <u,v>=\int_{\Omega} u(x)v(x) \ d\mu(x) $?
Now if $F\in \big(L^p(\Omega)^n\big)^*$. Does it work in the following way: $$F(u)= <u,F> = <(u_1,...,u_n),F>=<(u_1,...,u_n),v> \\= <(u_1,...,u_n),(v_1,...v_n)> =\sum_{i=1}^{n}<u_i,v_i>=\sum_{i=1}^{n} \int_{\Omega}u_i(x)v_i(x) \ d\mu(x)? $$
Yes, that's how it works. In general if $X$ is a normed vector space then $$(X^n)^*\cong(X^*)^n$$in a natural way.
To simplify the notation, note that by induction it's enough to show that $$(X\times Y)^*\cong X^*\times Y^*.$$Define $$i:X^*\times Y^*\to(X\times Y)^*$$by $\newcommand\ip[2]{\left\langle #1,#2\right\rangle}$ $$\ip{(x,y)}{i(x^*,y^*)}=\ip{x}{{x^*}}+\ip y{y^*}.$$It's easy to verify that in fact $i(x^*,y^*)\in(X\times Y)^*$. And given $\Lambda\in(X\times Y)^*$, define $x^*\in X^*$ and $y^*\in Y^*$ by $$\ip x{x^*}=\ip{(x,0)}{\Lambda},$$ $$\ip y{y^*}=\ip{(0,y)}{\Lambda};$$then it's easy to verify that $\Lambda=i(x^*,y^*)$ (because $(x,y)=(x,0)+(0,y)$.)
(The norm of $i(x^*,y^*)$ depends on what norm you use for $X\times Y$. For example if $||(x,y)||=||x||+||y||$ then $||i(x^*,y^*)||=\max||x^*||.||y^*||)$.)