How is the exterior power of a coherent sheaf is coherent?

So, I was reading Hans Grauert's and Reinhold Remmert's book "Theory of Stein Spaces" and in page 13, they define the exterior power $\bigwedge^{p} \mathcal{F}$ of a sheaf $\mathcal{F}$, and after that, they point out that if $\mathcal{F}$ is coherent, so will be $\bigwedge^{p} \mathcal{F}$ (all over a sheaf of rings $\mathcal{O}$). By the Three Lemma concerning coherent sheaves, since we have $\bigwedge^{p} \mathcal{F}:= \frac {\bigotimes^{p} \mathcal{F}}{\mathcal{M}}$, where $\mathcal{M}$ is a specific subsheaf of $\bigotimes^{p} \mathcal{F}$ described in the same page of the book,

$$ 0 \to \mathcal{M} \to \bigotimes^{p} \mathcal{F} \to \bigwedge^{p} \mathcal{F} \to 0 $$

is exact, which means, since $\bigotimes^{p} \mathcal{F}$ is coherent, that $\mathcal{M}$ is coherent if and only if $\bigwedge^{p} \mathcal{F}$ is coherent. And then it is necessary and sufficient to show that $\mathcal{M}$ is of finite type, and currently I don't know how to do that.

I appreciate any help and thank you in advance :)

Suppose $\mathcal{F}$ is locally generated by sections $s_{1},...,s_{n}$ over some open set. By definition, $\mathcal{M}$ is generated by sections of the form $a_{1} \otimes ... \otimes a_{p}$, with $a_{i}$ a section of $\mathcal{F}$, and with $a_{i} = a_{j}$ for some $i,j \in {1,...,p}$ such that $i \neq j$. Notice that if $k,l \in {1,...,n}$ with $k \neq l$, we have $(a_{1} \otimes ... \otimes a_{k} \otimes ... \otimes a_{l} \otimes ... \otimes a_{n}) + (a_{1} \otimes ... \otimes a_{l} \otimes ... \otimes a_{k} \otimes ... \otimes a_{n}) = (a_{1} \otimes ... \otimes a_{k} + a_{l} \otimes ... \otimes a_{k} + a_{l} \otimes ... \otimes a_{n}) - (a_{1} \otimes ... \otimes a_{k} \otimes ... \otimes a_{k} \otimes ... \otimes a_{n}) - (a_{1} \otimes ... \otimes a_{l} \otimes ... \otimes a_{l} \otimes ... \otimes a_{n}) \in \mathcal{M}$.

Having noted this, we now easily show that the set of sections $s_{i_1} \otimes ... \otimes s_{i_p}$, with $i:[1,...,p] \to [1,...,n]$ not injective (type $(1)$) and $(s_{i_{1}} \otimes ... \otimes s_{i_{k}} \otimes ... \otimes s_{i_{l}} \otimes ... \otimes s_{i_{n}}) + (s_{i_{1}} \otimes ... \otimes s_{i_{l}} \otimes ... \otimes s_{i_{k}} \otimes ... \otimes s_{i_{n}})$, with $i$ an arbitrary function as above, injective or not (type $(2)$).

Then, on staks, if we have $a_{j} = \sum \limits_{i_j = 1}^{n} c_{i_{j}}^{(j)} s_{i_j}$ with $j = 1,...,p$, then on stalks: $a_{1} \otimes ... \otimes a_{p} = \sum \limits_{i_1,...,i_p = 1}^{n} c_{i_1}^{(1)}...c_{i_p}^{(p)} s_{i_{1}} \otimes ... \otimes s_{i_{p}}$.

Since, for each set of fixed elements $i_{1},...,i_{p}$, we can arrange the elements in $p!$ ways. Since this discussion is only relevant for $p \geq 2$, we can arrange the sum in two parts: the first part as the sum of terms of type $(1)$ and the second part can be described as a sum of type $(2)$ terms, since $p!$ is even.

This shows $\mathcal{M}$ is of finite type, which is what we wanted.