I am reading "Calculus on Manifolds" by Michael Spivak.
The author wrote as follows:
Since each $\operatorname{Alt}(\varphi_{i_1}\otimes\cdots\otimes\varphi_{i_k})$ is a constant times one of the $\varphi_{i_1}\wedge\cdots\wedge\varphi_{i_k}$, these elements span $\Lambda^k(V)$.
I wonder why this holds.
My proof of the above fact:
Let $\omega_1,\dots,\omega_n\in\Lambda^1(V)$.
By the definition of the wedge product, $$\omega_1\wedge\omega_2=\frac{(1+1)!}{1!1!}\operatorname{Alt}(\omega_1\otimes\omega_2).$$
So, $\operatorname{Alt}(\omega_1\otimes\omega_2)=\frac{1}{2}(\omega_1\wedge\omega_2)$.
By Theorem 4-4(3) in the book, $$\omega_1\wedge\omega_2\wedge\omega_3=\frac{(1+1+1)!}{1!1!1!}\operatorname{Alt}(\omega_1\otimes\omega_2\otimes\omega_3).$$
So, $\operatorname{Alt}(\omega_1\otimes\omega_2\otimes\omega_3)=\frac{1}{6}(\omega_1\wedge\omega_2\wedge\omega_3)$.
By Theorem 4-4(2), $\operatorname{Alt}(\operatorname{Alt}(\omega\otimes\eta)\otimes\theta)=\operatorname{Alt}(\omega\otimes\eta\otimes\theta)$.
$$\operatorname{Alt}(\omega_1\otimes\omega_2\otimes\omega_3\otimes\omega_4)
=\operatorname{Alt}(\omega_1\otimes\omega_2\otimes(\omega_3\otimes\omega_4))
=\operatorname{Alt}(\operatorname{Alt}(\omega_1\otimes\omega_2)\otimes(\omega_3\otimes\omega_4)) \\
=\operatorname{Alt}\left(\frac{1}{2}(\omega_1\wedge\omega_2)\otimes\omega_3\otimes\omega_4\right)
=\frac{1}{12}(\omega_1\wedge\omega_2\wedge\omega_3\wedge\omega_4).$$
Let $n\geq 3$.
Similarly, $\operatorname{Alt}(\omega_1\otimes\cdots\otimes\omega_n)=\frac{1}{6}\left(\frac{1}{2}\right)^{n-3}(\omega_1\wedge\cdots\wedge\omega_n)$.
So, $\operatorname{Alt}(\varphi_{i_1}\otimes\cdots\otimes\varphi_{i_k})=\frac{1}{6}\left(\frac{1}{2}\right)^{k-3}(\varphi_{i_1}\wedge\cdots\wedge\varphi_{i_k})$ if $k\geq 3$.
Is my proof ok?
If my proof is not ok, please tell me a proof.
If my proof is ok, please tell me a better standard proof.
Your idea could be a start, but your constants are wrong.
You have showed that if each $\omega_i \in \Lambda^1(V)$, then
$$ \operatorname{Alt}(\omega_1 \otimes \omega_2) = \frac 12 (\omega_1 \wedge \omega_2) \tag{1}$$ $$ \operatorname{Alt}(\omega_1 \otimes \omega_2 \otimes \omega_3) = \frac 16 (\omega_1 \wedge \omega_2 \wedge \omega_3) \tag{2}$$
But then the first error is at
$$ \operatorname{Alt}\left(\frac 12 (\omega_1 \wedge \omega_2) \otimes \omega_3 \otimes \omega_4\right) = \frac{1}{12}(\omega_1 \wedge \omega_2 \wedge \omega_3 \wedge \omega_4) $$
You can't apply equation $(2)$ here since $\omega_1 \wedge \omega_2 \in \Lambda^2(V)$ is not an element of $\Lambda^1(V)$. Applying Spivak's Theorem 4-4(3) again, we instead have
$$ \begin{align*} \frac{(2+1+1)!}{(2!)(1!)(1!)} \operatorname{Alt}\left(\frac 12 (\omega_1 \wedge \omega_2) \otimes \omega_3 \otimes \omega_4\right) &= \frac 12(\omega_1 \wedge \omega_2 \wedge \omega_3 \wedge \omega_4) \\ \operatorname{Alt}\left(\omega_1 \otimes \omega_2 \otimes \omega_3 \otimes \omega_4\right) &= \frac{1}{24}(\omega_1 \wedge \omega_2 \wedge \omega_3 \wedge \omega_4) \tag{3} \end{align*} $$
$n$ isn't a good choice for your general formula about $\operatorname{Alt}(\omega_1 \otimes \cdots \otimes \omega_n)$ since $n$ already is used as the dimension of vector space $V$. Rather than just "Similarly" stating the general conclusion, a formal induction argument is better.
Rephrasing the essential ideas of your proof, I'd write:
(Or, both Spivak's 4-4(3) and this result (4) are special cases of a more general lemma not much harder to prove: If $\omega_i \in \Lambda^{j_i}$ for $i \in 1,\ldots,k$, then) $$ \operatorname{Alt}(\omega_1 \otimes \cdots \otimes \omega_k) = \frac{\prod_{i=1}^k j_i!}{\left(\sum_{i=1}^k j_i\right)!}(\omega_1 \wedge \cdots \wedge \omega_k) \tag{5} $$
However, statement (4) is not enough to finish the proof of Spivak's theorem 4-5! Note that in the basis decomposition
$$ \omega = \sum_{i_1,\ldots,i_k} a_{i_1,\ldots,i_k} \varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k} $$
the indices $i_1,\ldots,i_k$ may be any of the $n^k$ sequences of indices from $1$ to $n$, in any order and possibly with repeats. But the set of wedge products $\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k}$ we claim is a basis of $\Lambda^k(V)$ only uses those ${n \choose k}$ sequences of indices satisfying $1 \leq i_1 < i_2 < \cdots < i_k \leq n$.
But a rearrangment of index orders is involved in the definition of $\operatorname{Alt}$. So we have for any $w_1, \ldots, w_k \in V$,
$$ \begin{align*} \operatorname{Alt}(\varphi_{i_{\tau(1)}} \otimes \cdots \otimes \varphi_{i_{\tau(k)}})(w_1,\ldots,w_k) &= \sum_{\sigma \in S_k} \operatorname{sgn} \sigma \cdot (\varphi_{i_{\tau(1)}} \otimes \cdots \otimes \varphi_{i_{\tau(k)}})(w_{\sigma(1)},\ldots,w_{\sigma(2)}) \\ &= \sum_{\sigma \in S_k} \operatorname{sgn} \sigma \cdot (\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k})(w_{\tau^{-1}\sigma(1)}, \ldots, w_{\tau^{-1}\sigma(k)}) \\ &= \sum_{\sigma' \in S_k} \operatorname{sgn}(\tau \sigma') \cdot (\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k})(w_{\sigma'(1)}, \ldots, w_{\sigma'(k)}) \\ &= \operatorname{sgn}\tau \cdot \operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k})(w_1,\ldots,w_k) \\ \operatorname{Alt}(\varphi_{i_{\tau(1)}} \otimes \cdots \otimes \varphi_{i_{\tau(k)}}) &= \operatorname{sgn}\tau \cdot \operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k}) \tag{6} \end{align*} $$
If any index appears more than once in the sequence $i_1,\ldots,i_k$, say $i_a=i_b$ with $a \neq b$, and if $\tau = (ab)$ is the pair swap permutation, then $i_{\tau(j)} = i_j$ for every $j$, and $\operatorname{sgn} \tau = -1$. So equation (6) gives
$$ \operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k}) = - \operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k}) $$
and therefore $\operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k}) = 0$. (It's technically true this is a constant times any one of the alternating wedge product basis elements, where that constant is zero. But this case could have been explained better!)
Otherwise, $i_1,\ldots,i_k$ are $k$ distinct elements of $\{1,\ldots,n\}$. Therefore they can be rearranged by some permutation $\tau$ so that $1 \leq \tau(i_1) < \cdots < \tau(i_k) \leq n$. And combining equations (4) and (6),
$$ \begin{align*} \operatorname{sgn} \tau \cdot \operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k}) &= \operatorname{Alt}(\varphi_{i_{\tau(1)}} \otimes \cdots \otimes \varphi_{i_{\tau(k)}}) \\ \operatorname{Alt}(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k}) &= \frac{\operatorname{sgn} \tau}{k!} \varphi_{i_{\tau(1)}} \wedge \cdots \wedge \varphi_{i_{\tau(k)}} \end{align*} $$
This proves that the set $\{\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k} \mid 1 \leq i_1 < \cdots < i_k \leq n\}$ spans $\Lambda^k(V)$. To prove it's a basis, it remains to show these wedge products are linearly independent. For this I'd suggest the lemma:
If $1 \leq i_1 < \cdots < i_k \leq n$ and $1 \leq j_1 < \cdots < j_k \leq n$, then $$ (\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_k})(v_{j_1}, \ldots, v_{j_k}) = \begin{cases} 1 & \mathrm{if\ } \forall m (i_m = j_m)\\ 0 & \mathrm{if\ } \exists m (i_m \neq j_m) \end{cases} $$