Respuesta :
Answer:
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
Answer:
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