This problem is already solved in an elegant way here:
Product of spheres embeds in Euclidean space of 1 dimension higher
But I was trying to use a different approach:
I'm using induction, it's clear that $\mathbb{S}^1$ is a embebed submanifold of $\mathbb{R}^2$ and in the same way $\mathbb{S}^n$ is an embebed submanifold of $\mathbb{R}^{n+1}$.
For the inductive step let's take $\mathbb{S}^{n_1}\times ...\times \mathbb{S}^{n_k}$ embebed submanifold of $\mathbb{R}^{n_1+...+n_k+1}$ and let's prove that $\mathbb{S}^{n_1}\times ...\times \mathbb{S}^{n_k}\times \mathbb{S}^{k}$ it's a embebed submanifold of $\mathbb{R}^{n_1+...+n_k+k+1}$.
To do it, I think we can consider that $\mathbb{S}^{k}\times M$ where $M=\mathbb{S}^{n_1}\times ...\times \mathbb{S}^{n_k}$ is isomorphic to $SO(k+1)\times M$, is that true, right? So I can think $SO(k+1)\times M$ as $M$ is moving arroung the k+1 axis, leaving $k+1$ components with the same position, but I dont know how to express this idea formally, can someone help me please?
Denote $S^{n_1}\times\dots\times S^{n_k}=M\subset \mathbb{R}^n=\mathbb{R}^{n_1+\dots n_k+1}$ as the embedded manifold of the inductive step. We want to prove that $S^k\times M$ embeds in $\mathbb{R}^{n+k}$. Since $M$ has codimension $k$ in $\mathbb{R}^{n+k-1}$, for $\epsilon$ small enough the $\epsilon$-tubular neigbhourhood of $M\subset\mathbb{R}^{n+k-1}$ is a product bundle $M'=M\times D^k_{\epsilon}$. If you embed $M'$ in $\mathbb{R}^{n+k}$ say at cooridnate $x_{n+k}=0$ and $M$ at coordinate $x_{n+k}=1$, then for every $x\in M$ you can connect the boundary of $M'$ around $x$ to $x$ in $M$, this amouonts to do a pointwise suspension to the bundle, i.e. you take the boundary of the cone $\partial((D^k_\epsilon\times I)/\sim)\cong S^k$, where $\sim$ collapse $D^k_\epsilon\times {1}$ to a point.
I think in this way you produced a manifold that is homeomorphic to $M\times S^k$ embedded in $\mathbb{R}^{n+k}$