# Why $W^{1,n}\subset\subset L^{q}$ for $q\in [n,\infty)$?

I am reading the book named "Functional Analysis, Sobolev Spaces, and Partial differential Equations" by Brezis. In the book, the Rellich-Kondrachov theorem is stated as follows

Suppose that $$\Omega$$ is bounded and of class $$C^1$$ in $$\mathbb{R}^n$$. Then we have the following compact injections:
\begin{align} W^{1,p}(\Omega)&\subset L^q(\Omega)& \forall q&\in [1,np/(n-p)) \quad pn. \end{align}

In the book, the author states that the case $$p=n$$ can be reduced to the case $$p. I am confused of it, can you give me some hints?

Brezis, Haim, Functional analysis, Sobolev spaces and partial differential equations, Universitext. New York, NY: Springer (ISBN 978-0-387-70913-0/pbk; 978-0-387-70914-7/ebook). xiii, 599 p. (2011). ZBL1220.46002.

This is simply because on a bounded domain, if $$t>0$$ then $$L^{p+t}\subset L^p$$.
So if $$f\in W^{1,n}$$, then $$f\in W^{1,p}$$ for all $$p. For each such $$p$$, applying the $$p result shows that $$W^{1,n}\subset L^q$$ for all $$q\in [1, np/(n-p))$$. (Also recall bounded composed with compact is compact)
But $$np/(n-p)$$ is arbitrarily large by taking $$p\to n$$, giving the full range $$[1,\infty)$$. (this is the same as saying $$[n,\infty)$$ because of the same $$L^{p+t}\subset L^p$$ reason as above)