# which number is bigger (sqrt(8)^sqrt(7)) or (sqrt(7)^(sqrt(8))can we extend that for (sqrt(n+1)^sqrt(n)) or (sqrt(n)^(sqrt(n+1))

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We know that

(n+1)^m < n^m+n^n+{nC2*n^n-2+ (n-1) lesser terms}

<2n^m+{nC2*n^(n-2)}n

<2n^m+(n^(n+1)-n^(n))/2

=n^n{(n+3)/2}<n^(n+1), for all n>(n+3)/2 or n>3.

Thus n^n+1 > (n+1)^n=>

**n^((n+1)^(1/2) > (n+1)^n^(1/2).-----------(1)**

Using the result, n=7 and n+1=8,

7^8^(1/2) > 8^&^(1/2) is established. Take the square root and rewriting , we get,

sqrt of {7^(sqrt8)} > sqrt{8^(sqrt7)} or

{sqrt7}^(sqrt8) > {sqrt8}^(sqrt7).

Taking square on both sides of the relation at flag(1) we arrive at the second part which generalises the result after changing the sides.

**(sqrt(n+1)^sqrt(n))** < **(sqrt(n)^(sqrt(n+1))**

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