In what follows, $U$ is a complex vector space. I have the following abstract definition.
Definition: Let $\wedge^{n}: \overbrace{U \times \cdots \times U}^{\text{$n$ times}} \to V$ be an alternating $p$-linear map. We say that $\wedge^{p}$ satisfies the universal property if given another alternating $p$-linear map $f: U \times \cdots \times U \to W$, then there exists a unique map $h: V \to W$ such that $f = h\circ \wedge^{n}$.
Now, suppose $U$ is finite dimensional. Consider $V = \text{Alt}^{n}(U)$ the space of alternating tensors and let: $$\wedge^{n}(v_{1},...,v_{n}):= \frac{1}{n!}\sum_{\sigma \in S_{n}}\text{sign}(\sigma)v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)}$$
This map is clearly alternating an $p$-linear. My question is: does it also satisfy the universal property? I couldn't prove it myself.
$ \newcommand\Tensor{\mathop{\textstyle\bigotimes}} \newcommand\TensPow[1]{\mathop{\textstyle\bigotimes^{#1}}} \newcommand\Alt{\mathop{\mathrm{Alt}}\nolimits} \newcommand\alt{\mathop{\mathrm{alt}}} $ Let $\TensPow nU$ denote the $n^{\text{th}}$ tensor power of $U$. Define $$ \alt : \TensPow nU \to \Alt^n(U),\quad \alt(u_1\otimes\dotsb\otimes u_n) = \frac1{n!}\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)u_{\sigma(1)}\otimes\dotsb\otimes u_{\sigma(n)} $$ using the universal property of $\TensPow nU$. Also define $$ i : U^n \to \TensPow nU,\quad i(u_1,\dotsc,u_n) = u_1\otimes\dotsb\otimes u_n $$ so that ${\wedge^n} : U^n \to \Alt^n(U)$ is ${\wedge^n} = {\alt}\circ i$. Again by the universal property of the tensor power $\TensPow nU$, any alternating multilinear $f : U^n \to W$ extends uniquely to $f' : \TensPow nU \to W$ such that $f = f'\circ i$. Restrict $f'$ to $\Alt^n(U)$ to get a map $\Alt^n(U) \to W$. Because $f$ is alternating we find that $f' = f'\circ{\alt}$; hence $$ f = f'\circ i = f'\circ{\alt}\circ i = f'\circ{\wedge^n}. $$
We now wish to show that $f'$ is unique. To that end, suppose we have $f = h\circ{\wedge^n} = h\circ{\alt}\circ i$ for some $h$. By the universal property of $\TensPow nU$ it must be that $$ h\circ{\alt} = f' = f'\circ{\alt}. $$ But then $h = f'$ since $\alt$ is surjective.