Let $H$ be a Hilbert space and for $a,b\in H$ denote by $a\otimes b$ the tensor product $a,b$ which is realized as a multilinear map on $H^2$, i.e. $(a\otimes b)(\lambda v, u) = \lambda^*(a\otimes b)(v,u)$ and $(a\otimes b)(v + u,w) = (a\otimes b)(v,w) + (a\otimes b)(u,w)$ in both components. $a\otimes b$ has an explicit representation with $(a\otimes b)(v,u) := \langle a,v\rangle \langle b,u\rangle$. Norm of $a\otimes b$ is given by
$$\|a\otimes b\| = \sup_{v\in H^2:\|v\|\leq 1}|(a\otimes b)(v_1,v_2)|$$
A proof I am reading makes the claim w/o a proof that if $e_1,\dots,e_n$ are Hilbert basis elements of $H$ and $a\in H$, then
$$\left\|a\otimes\bigotimes_{i=1}^ne_i\right\|^2=\|a\|^2\left\|\bigotimes_{i=1}^ne_i\right\|^2$$
I don't quite understand this claim in the case that $a\neq 0$: For $v\in H^{n+1}$ we necessarily need to have $\|v_1\| > 0$. Therefore the product $\prod_{i=1}^{n}\langle e_i,v_{i+1}\rangle$ has necessarily "less mass" in the $v_{i+1}$s, since in $\left\|a\otimes\bigotimes_{i=1}^ne_i\right\|$'s norm we are taking the supremum over all $v\in H^{n+1}$ with $1\geq \|v\| = \sqrt{\sum_{i=1}^{n+1}\|v_i\|^2}$.
Am I missing something or is the claim false?
If $\|(v_1,v_2)\|\leq1$, this is $\|v_1\|^2+\|v_2\|^2\leq1$. Then \begin{align}\def\abajo{\\[0.3cm]} |(a\otimes b)(v_1,v_2)| &=|\langle a,v_1\rangle\langle b,v_2\rangle|\leq\|a\|\,\|b\|\,\|v_1\|\,\|v_2\|\abajo &\leq\|a\|\,\|b\|\,\Big(\frac12\,\|v_1\|^2+\|v_2\|^2\Big)\abajo &\leq\frac12\,\|a\|\,\|b\|. \end{align} Taking $v_1=\frac a{\sqrt2\,\|a\|}$, $v_2=\frac b{\sqrt 2\,\|b\|}$, $$ a\otimes b(v_1,v_2)=\frac12\,\|a\|\,\|b\|. $$ So $$\tag1 \|a\otimes b\|=\frac12\,\|a\|\,\|b\|. $$ This makes the equality you want impossible. But, more importantly, the norm you defined is not a Hilbert space norm. Indeed, if $\{e_n\}$ are orthonormal in $H$ consider $\|e_1\otimes e_1+e_2\otimes e_2\| $. For any $(v_1,v_2)$ you have $$ |(e_1\otimes e_1+e_2\otimes e_2)(v_1,v_2)| =|\langle e_1,v_1\rangle\langle e_1,v_2\rangle+\langle e_2,v_1\rangle\langle e_2,v_2\rangle|. $$ It is easy to see from this that $\|e_1\otimes e_1\pm e_2\otimes e_2\|=\frac12$. So $$ \|e_1\otimes e_1+e_2\otimes e_2\|^2+\|e_1\otimes e_1-e_2\otimes e_2\|^2 =\frac12\ne 1=2\|e_1\otimes e_1\|^2+2\|e_2\otimes e_2\|^2. $$ This shows that the norm fails the parallelogram identity and thus is not a Hilbert space norm.
The usual way the norm is defined on the tensor product of Hilbert spaces is by defining first the inner product. This is done by $$ \langle a\otimes b,c\otimes d\rangle=\langle a,c\rangle\langle b,d\rangle $$ and extending by (sesqui)linearity. This gives you automatically that for elementary tensors you have $$ \|a\otimes b\|=\langle a\otimes b,a\otimes b\rangle=\langle a,a\rangle^{1/2}\langle b,b\rangle^{1/2}=\|a\|\,\|b\|. $$