An elementary proof was shown to me by <A HREF="https://mathscinet.ams.org/mathscinet/search/author.html?mrauthid=210365"><FONT FACE="Arial">**Jan Malý**</FONT></A><FONT FACE="Arial">. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.




> **Theorem.** For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap
 C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.}
 $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into
> the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of
> $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

> **Corollary.** If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ 
 E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad
 \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

**[1] P. Goldstein, P.  Hajłasz,** 
<A HREF="https://arxiv.org/abs/1812.11888"><FONT FACE="Arial">Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$</FONT></A><FONT FACE="Arial">. (Preprint 2018), arXiv:1812.11888.