Assume $f\in C^1_c(\mathbb{R}^n), f\geq 0$. Define $$ f_t:=\min\{t,f\},\quad \chi(t):=\left(\int_{\mathbb{R}^n} f_t^{\frac{n}{n-1}}dx\right)^{\frac{n-1}{n}}. $$ Then $\chi$ is nondecreasing on $(0,\infty)$ and $$ \lim_{t\to\infty} \chi(t) = \left(\int_{\mathbb{R}^n} |f|^{\frac{n}{n-1}}dx\right)^{\frac{n-1}{n}}.\tag{$\star$} $$
I have already proved that $\chi$ is nondecreasing in $(0,\infty)$ and since $\chi$ is a bounded monotone function the limit $\lim\limits_{t\to\infty} \chi(t)$ exists, but why is it equal to $(\star)$? Moreover, $f=|f|$ since $f\geq 0$. So actually $$ \lim_{t\to\infty} \chi(t) = \left(\int_{\mathbb{R}^n} |f|^{\frac{n}{n-1}}dx\right)^{\frac{n-1}{n}} = \left(\int_{\mathbb{R}^n} f^{\frac{n}{n-1}}dx\right)^{\frac{n-1}{n}}. $$ I think it is possible to solve this problem by the Monotone Convergence Theorem because $0\leq f_t\leq f_{t+1}$ for every $t\in (0,\infty)$. But I do not know if it's true that $$ \lim_{t\to\infty} f_t = \lim_{t\to\infty} \min\{t,f\} = f. $$
For any $x\in \Bbb R^n$, $f(x)<M$ for some $M>0$ since $f(x)$ is finite. For all $t>M$, you get that $$ f_t(x) = f(x) $$ so $\lim_{t\to \infty} f_t(x) = f(x)$. This shows that $f_t \to f$ pointwise everywhere.