Where do you get these problems?

This is a great problem!

Is this problem written wrong? Shouldn't the $t$ not be in the exponent?

Does anyone have solution for this?

This problem is just an obvious statement obfuscated by complicated notation. iff direction: If $n$ is prime, then $\dfrac{(u!)^r}{n}$ will not be an integer for $u < n$, but it will be a multiple of $2$ for $u\ge n$. Thus, for $u<n$, since $\left|\cos\,\frac{(u!)^{r} \pi}{n} \right | <1$, $$\lim\limits_{r\to\infty}\lim\limits_{t\to \infty}\left(\cos\,\frac{(u!)^{r} \pi}{n} \right)^{2t} =0 \implies \lim\limits_{r\to\infty}\lim\limits_{t\to \infty}1-\left(\cos\,\frac{(u!)^{r} \pi}{n} \right)^{2t} =1$$Meanwhile for $u\ge n$, $\cos\,\frac{(u!)^{r} \pi}{n}=1$ so overall $$\lim_{r \rightarrow\infty}\,\lim_{s \rightarrow\infty}\,\lim_{t \rightarrow \infty}\,\sum_{u=0}^{s}\left(1-\left(\cos\,\frac{(u!)^{r} \pi}{n} \right)^{2t} \right)=\sum_{u=0}^{n-1} 1 = n$$if direction: If $n$ is not prime, then there exists $a>b>1$ such that $n=ab$. Thus, for $u \ge a$, $\dfrac{(u!)^r}{n}$ will be a multiple of 2, so $\cos\,\frac{(u!)^{r} \pi}{n}=1$. Thus, $$\lim_{r \rightarrow\infty}\,\lim_{s \rightarrow\infty}\,\lim_{t \rightarrow \infty}\,\sum_{u=0}^{s}\left(1-\left(\cos\,\frac{(u!)^{r} \pi}{n} \right)^{2t} \right) \le \sum_{u=0}^{a-1} 1 = a < n$$Both directions are proved so the problem is finished. $\Box$