Why is this line bundle clearly ample? Reading a book I encountered the following claim, which I don't understand. Let $X$ be a smooth projective curve over $\mathbb{C}$, and $q\in X$ a rational point.
Denote by $\pi_i: X^n\to X$ the $i$-th projection of the cartesian $n$-product of the curve onto $X$ itself. The claim is that

The line bundle $\bigotimes_{i=1}^n \pi_i^* \mathcal{O}_X(q) $ is clearly ample.

Could you point me in the right direction please? Is there a specific criterion for ampleness I should immediately see it's satisfied?
I know that, being each $\pi_i$ finite and surjective, every $\pi_i^*\mathcal{O}_X(q)$ is ample because each $\mathcal{O}_X(q)$ is. But how can one conclude from here that the tensor product of them is?
PS: Is there any way to see this geometrically?
 A: Geometrically the tensor product $L$ is ample because a sufficiently high power $L^N$ is the tensor product of pullbacks of very ample bundles on $X$.
In more detail, say the $N$-fold tensor power $\mathcal{O}_X(q)^N$ is very ample, and that the sections $s_1, \dots, s_m$ projectively embed $X$. For each $i = 1, \dots, n$ and $j = 1, \dots, m$, let $s_{i,j} = \pi_i^* s_j$ denote the pullback of $s_j$ by projection to the $i$th factor. The $m^n$ sections $s_{1,f(1)} \otimes \dots \otimes s_{n,f(n)}$ of $L^N$ (as $f$ ranges over all mappings from $\{1, \dots, n\}$ to $\{1, \dots, m\}$) embed the $n$-fold product $X \times \dots \times X$.
FWIW, sections of the pullback $\pi_i^* \mathcal{O}_X(q)$ are "non-constant only in the $i$th factor", so it appears they don't separate points except (possibly) in the $i$th factor. :)
A: The result you want is the following.


Hartshorne II Ex. 7.5 (c):  If $\mathcal{L}, \mathcal{M}$ are two ample line bundles on a Noetherian scheme $X$, then $\mathcal{L}\otimes \mathcal{M}$ is ample. 


Proof: As $\mathcal{M}$ is ample there is some $m > 0$ for which $\mathcal{M}^{\otimes n}$ is generated by its global sections. Now $\mathcal{L}^{\otimes m}$ is still ample  in view of Proposition 7.5. Thus $\mathcal{M}^{\otimes m} \otimes \mathcal{L}^{\otimes m} = (\mathcal{M} \otimes \mathcal{L})^{\otimes m}$ is ample by part (a). By Proposition 7.5 again we conclude $\mathcal{M} \otimes \mathcal{L}$ is ample. 


Exercise: Prove part (a) of the exercise above, namely if $\mathcal{L}$ is ample and $\mathcal{M}$ is globally generated then $\mathcal{L}\otimes \mathcal{M}$ is ample.


