Sobolev embedding when $p=n$: $W^{1,p}(\mathbb{R}^{n}) \hookrightarrow L^{q}(\mathbb{R}^{n})$ for $q: p \leqslant q < \infty$

In class, aside from the standard Gagliardo-Nirenberg-Sobolev and Morrey inequalities, my professor also covered the case when $$p=n$$. In particular, if $$p=n$$, then $$W^{1,p}(\mathbb{R}^{n}) \hookrightarrow L^{q}(\mathbb{R}^{n})$$ for all $$q$$ such that $$p \leqslant q < \infty$$. We did not prove this case in class.

I can see that this would be true trivially whenever $$q=p$$: $$||u||_{L^{p}} \leqslant ||u||_{W^{1,p}}$$. If it was true for $$q=\infty$$, then I could interpolate to immediately get the result for $$p < q < \infty$$. But I know it's definitely not true when $$q=\infty$$ (consider $$\log\log(1+|x|^{-1})$$, which is in $$W^{1,n}(B(0,1))$$ but not $$L^{\infty}(B(0,1))$$).

Can anyone provide some guidance on this for me? At least if $$u \in C_{c}^{\infty}(\mathbb{R}^{n})$$ and $$u\neq 0$$, if I define $$u_{\lambda}(x) = u(\lambda x)$$ and assume that the inequality $$||u_{\lambda}||_{L^{q}} \leqslant C||u_{\lambda}||_{W^{1,n}}$$ holds, I cannot get a contradiction anymore as in the case when $$1 \leqslant p < n$$, where the contradiction is that $$u=0$$ a.e. by sending $$\lambda$$ to either $$0$$ or $$\infty$$.

Edit: I found that the proof is actually done by Brezis (Corollary 9.11). I may transcribe the answer later if I have time.

On a bounded domain $$\Omega$$, one can do the following: For $$q \in [p,\infty)$$ you can find $$\tilde p < p = n$$ with $$\frac1q = \frac1{\tilde p} - \frac1n.$$ Now, you can use the 'standard' Sobolev embedding for $$\tilde p < n$$ to obtain $$W^{1,p}(\Omega) \hookrightarrow W^{1,\tilde p}(\Omega) \hookrightarrow L^q(\Omega).$$
Actually, in the limiting case for $$p>1$$ one can prove a stronger result, known as Trudinger inequality $$W^{1,p}(\Omega) \hookrightarrow L^\varphi(\Omega)$$ where $$\varphi$$ is defined as follows $$\varphi(t) = \exp\left(t^{\frac{p}{p-1}}\right) - 1.$$