Given the definition of a Toric variety, I can see that projective space contains a torus by:
$({\mathbb{C}^*})^{n}\cong\{[1:t_1:...:t_n]:t_i\in {\mathbb{C}^*}, i =1,...,n\}\subset\mathbb{P}^{n}$
but I am failing to see how this is zariski open, indeed most examples simply give the isomorphism to a torus and then don't show it is zariski open. Any help would be appreciated!!
Also, why is the above torus used rather than: $({\mathbb{C}^*})^{n+1}\cong\{[t_0:t_1:...:t_n]:t_i\in {\mathbb{C}^*}, i =0,...,n\}\subset\mathbb{P}^{n}$
First of all, the $(n+1)$-torus you describe isn't actually an $(n+1)$-torus. Since projective coordinates are only defined up to scalar multiplication, the point $[t_0 : t_1 : \cdots : t_n]$ is equal to $[1 : \frac{t_1}{t_0} : \cdots : \frac{t_n}{t_0}]$, which makes sense because $t_0 \neq 0$. So every point described by your second equation is also described by the first.
As far as openness, recall that a subset of $\mathbb P^n$ is Zariski-closed if and only if it is the vanishing locus of some set of homogeneous polynomials in the coordinate variables $x_0, \dots, x_n$; a subset is open if it is the complement of a closed set. In this case, your torus is the complement of the set $$ \{ [x_0 : x_1 : \cdots : x_n]: x_0 x_1 \cdots x_n = 0 \}; $$ that is, the vanishing locus of the homogeneous polynomial $x_0 x_1 \cdots x_n$.