Product of spheres embeds in Euclidean space of 1 dimension higher

• Note first that $\mathbb{R}\times\mathbb{S}^n$ smoothly embeds in $\mathbb{R}^{n+1}$ for each $n$, via $(t,\textbf{p})\mapsto e^t\textbf{p}$.
• Taking the Cartesian product with $\mathbb{R}^{m-1}$, we find that $\mathbb{R}^m\times\mathbb{S}^n$ smoothly embeds in $\mathbb{R}^{m}\times\mathbb{R}^n$ for each $m$ and $n$.
• By induction, it follows that $\mathbb{R}\times\prod_{i=1}^m \mathbb{S}^{n_i}$ smoothly embeds in a Euclidean space of dimension $1+\sum_{i=1}^m n_i$.

The desired statement follows.

EDIT: I cannot delete this post as it's been already accepted :(

The proof is carried out by induction on $m.$

$m=1$ is trivial by choosing coordinates $(x^{(1)}_0,x^{(1)}_1,...,x^{(1)}_{n_1})$ where $\sum_{j=0}^{n_1} (x_j^{(1)})^2=1$, so let $m=2,$ then similarly $S^{n_1} \times S^{n_2}$ is embedded in $\mathbb{R}^{n_1+n_2+2}$ which also lies in the hypersurface $H$ with equation $\sum_{j=0}^{n_1} (x_j^{(1)})^2+\sum_{j=0}^{n_2} (x_j^{(2)})^2=2.$ In fact, $H$ is diffeomorphic to $S^{n_1+n_2+1}.$ Now, the embedding is missing at least a point, for example $p=(0,...,0,1,1) \in H,$ so by stereographic projection $S^{n_1+n_2+1} \setminus \{p\}$ is diffeomorphic to $\mathbb{R}^{n_1+n_2+1}.$

Suppose that the assertion holds for $<m,$ then $(S^{n_1}\times...\times S^{n_{m-1}}) \times S^{n_m}$ is embedded diffeomorphically in $\mathbb{R}^{n_1+...+n_{m-1}+1} \times \mathbb{R}^{n_m+1}\cong \mathbb{R}^{n_1+...+n_{m}+2}$ hence following the same argument you can reduce the dimension by $1,$ so the result.