用了快速幂、一层 for 循环，同时用了快读，时间复杂度应该是 $O(n \log n)$ 的，但最后两个测试点 TLE 了。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const LL MOD = 998244353;
int read()
{
	int x = 0, w = 1;
	char ch = 0;
	while(ch < '0' || ch > '9')
	{
		if(ch == '-') w = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9')
	{
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x * w;
}
long long fast_pow(long long a, long long b, long long p)
{
	LL res = 1;
	while(b)
	{
		if(b & 1) res = res * a % p;
		a = a * a % p;
		b >>= 1;
	}
	return res;
}
int main()
{
	int n;
	n = read();
	//scanf("%d", &n);
	LL ans = 1;
	for(int k = 1; k <= n; k ++)
	{
		ans = ans * fast_pow((LL)k, 1ll * k * (n - k + 2) * (n - k + 1) / 2, MOD) % MOD;
	}
	printf("%lld\n", ans);
	return 0;
}