Prove by induction that n! ≤ n^n

for all n ∈ N.

n! ≤ n^n

Step-by-step explanation:

n! ≤ n^n

Proof

let n=1

1!=1=1^1=1

hence 1=1

when n=2

2!=1x2=2 and 2^2 = 2x2=4

hence 2≤4

when n=n+1, (n+1) !=n! (n+1) = (n+1) ^ (n+1) = (n+1) ^n x (n+1)

i. e. n! (n+1) = (n+1) ^nXn+1

Divide both sides by n+1

n! = (n+1) ^n

hence n! ≤ n^n

the equation given satisfies the given condition of n!<=n^n

Step-by-step explanation:

taking n=4 and n=2 and n=1

4! < = 4^4

4*3*2*1 < = 256

24 <256

2! < = 2^2

2*1 < = 4

2 < 4

1! < = 1^1

1*1 < = 1

1=1

hence proved