N + 1 power of N and (n + 1) power of N, compare the size

N + 1 power of N and (n + 1) power of N, compare the size

If 0 ≤ n < 3
The N + 1 power of n is less than the n power of (n + 1)
If n ≥ 3
N + 1 power of n ＞ (n + 1) power of n
If n ＜ 0 and the integer bit of the absolute value of n is even
The N + 1 power of n is less than the n power of (n + 1)
If n ＜ 0 and the integer part of the absolute value of n is odd (n ≠ - 1, because if n = - 1, the nth power of (n + 1) is the negative power of 0, that is, 1 / 0, 0 cannot be a divisor, so n ≠ - 1)
N + 1 power of n ＞ (n + 1) power of n
When n = 0
The N + 1 power of n is 0 to the 1 power of 0, the n power of (n + 1) is 0 to the 0 power of 1, and the 0 power of any number is 1

(n) The comparison of the power of (n + 1) and the power of (n + 1)

Do this problem with the limit to do, when n take a large value to compare the relationship between the two
limit (n^n+1) / (n+1)^n = (n^n * n) / (n+1)^n
n->infitiry
Since n is close to infinity, n is approximately equal to N + 1. Therefore, we can divide n ^ n and (n + 1) ^ n, so the limit becomes limit = n, and N is infinite, that is, the numerator is larger than the denominator
Although n ^ n + 1 and (n + 1) ^ n tend to infinity, n ^ n + 1 approaches infinity much faster than (n + 1) ^ n

Compare the N + 1 power of N and the n power of (n + 1)

Let's prove (n + 1) ^ n / N ^ n = 3)
（n+1)^n/n^n
=(1+1/n)^n
=1+C(n,1)*1/n+C(n,2)1/n^2+...