I am reading Calculus on Manifolds, and Spivak stated that a k-tensor $\omega \in J^{k}(V)$ is called alternating if $\omega(v_{1}, \dots,v_{i}, \dots, v_{j}, \dots, v_{k}) = -\omega(v_{1}, \dots,v_{j},\dots v_{i},\dots v_{k})$ for all $v_{1},\dots v_{k} \in V$
I think if $\omega \in J^{n+1}(V)$ where $dim(V) = n$, then $\omega = 0$. Is this true? If it is, is there a simple way to see this from the above definition?
I think this is true because we have $$ \omega = \sum_{i_{1},\dots i_{n+1}}a_{i_{1}, \dots ,i_{n+1}}\phi_{i_{1}}\otimes \phi_{i_{2}}\otimes\dots \otimes \phi_{i_{n+1}} $$, where $\phi_{1},\dots \phi_{n}$ is the dual basis of $V$,
We also have $$ Alt(\omega) = \omega = \sum_{i_{1},\dots i_{n+1}}a_{i_{1}, \dots ,i_{n+1}} Alt(\phi_{i_{1}}\otimes \phi_{i_{2}}\otimes\dots \otimes \phi_{i_{n+1}}) = \sum_{i_{1},\dots i_{n+1}}b_{i_{1}, \dots ,i_{n+1}} \phi_{i_{1}} \wedge \dots \wedge \phi_{i_{n+1}} $$ but as there exists $i_{s},i_{t}$, such that $\phi_{i_{s}} = \phi_{i_{t}}$, so $\phi_{i_{1}} \wedge\dots \wedge\phi_{i_{s}}\wedge\dots \wedge\phi_{i_{t}}\wedge\dots \wedge\phi_{i_{n+1}} = -\phi_{i_{1}} \wedge\dots \wedge\phi_{i_{t}}\wedge\dots \wedge\phi_{i_{s}}\wedge\dots \wedge\phi_{i_{n+1}}$ which implies $\phi_{i_{1}}\wedge\dots \wedge\phi_{i_{n+1}} = 0$, so $\omega = 0$
I am not sure if my argument is correct, and even if it is, is there a simpler way to show this?
Your argument is correct, I think the simplest way to see this as follows: First observe that the definition you give implies that $\omega$ vanishes if two of its entries are equal. This easily implies that $\omega$ vanishes whenever the vectors you insert are linearly dependent (since then you can write one of them as a linear combination of the others). Since any $n+1$ vectors in an $n$-dimensional vector space are linearly dependent, the claim follows.