Is $\mathbb{C}^n$ a C*-algebra? Hi I am new in learning C*-algebra. I know that under the usual $\mathbb{C}^n$-norm, $\mathbb{C}^n$ is a Banach Space. However, what will be an intuitive multiplication on $\mathbb{C}^n$ so that the usual $\mathbb{C}^n$-norm will be sub-multiplicative? In fact, I should prove that the usual $\mathbb{C}^n$-norm is a C* norm too. I was trying pointwise multiplication but I was stuck in the calculations. 
Edited(my calculations):
I used for $u=(u_1,u_2,...,u_n),v=(v_1,v_2,...,v_n)\in \mathbb{C}^n$, $uv= (u_1v_1,u_2v_2,...,u_nv_n)$. So I am trying to proceed in proving $\|uv\|_n \le \|u\|_n \|v\|_n$. In particular I am stuck in proving that $\|uv\|_n=\|(u_1v_1,u_2v_2,...,u_nv_n)\|_n \le \|u\|_n \|v\|_n$. What I mean is I don't know how to "split" $(u_1v_1,u_2v_2,...,u_nv_n)$ into relations of $(u_1,u_2,...,u_n)$ and $(v_1,v_2,...,v_n)$.
Another edit: By the answers below, $\mathbb{C}^n$ will be not a C*-algebra under the usual $\mathbb{C}^n$-norm. 
 A: You can write $(\Bbb C^n, \lVert\cdot\rVert_p)=L^p(\Bbb Z/n\Bbb Z) = L^p(G)$, the inner group is compact, hence when $p\ge 1$ you have a Banach space structure and the usual convolution on functions over a compact group give you an algebra structure, given as

$$(f\star g)(x)=\int_G f(y)g(x-y)\,dy.$$

Here "-" means the subtraction in the group structure on $G$. In your case integration is just a sum, so you might more easily recognize this as
$$(f\star g)(x)=\sum_{i=1}^n f(y_i)g(x-y_i)$$
with the sum over $y_i\in G$. If you're unfamiliar with this construction, it's analogous to the one done on $L^1(\Bbb R^d)$ and Wikipedia has a brief mentioning of the construction. The "unomodular" hypothesis they allude to is satisfied by compact abelian groups (it means there is a uniform measure, and for finite groups this is easily seen to be true by assigning each point measure $|G|^{-1}$.
Edit I got too caught up in just the Banach algebra structure, I left out the $C^*$ condition. For the $C^*$ structure, you have to fine tune a bit, using the infinity norm we see that $f\mapsto\overline{f}$ is the required $*$ operation. The norm is clearly preserved in this way since the sup of $|\overline{f}|$ doesn't change and $\lVert f^*f\rVert_\infty = \sup |f\overline{f}|=\sup |f|^2$ by positivity of the absolute value, this is clearly $(\sup |f|)^2$ which is the product of the two norms (this doesn't work for $p<\infty$ which you can easily check).
But it is important to note this does not work for $p=2$, which means your question fails for the usual norm induced by the Hermitian inner-product.
