0 to the power of 0

hi beary!

if you ask the calculator, it says it is zero.

however, the alternating sum of binomial coefficients from the n-th row of Pascal's triangle is what you obtain by expanding (1-1)^n using the binomial theorem, i.e., 0n.

 But the alternating sum of the entries of every row except the top row is 0, since 0^k=0 for all k greater than 1.

But the top row of Pascal's triangle contains a single 1, so its alternating sum is 1, which supports the notion that (1-1)^0=0^0 if it were defined, should be 1.

so therefore 0^0 is zero.

i hope this helped,

gavin

0⁰  is an indeterminant form...here's why ...

Note...if we let n be any real number...we have

0⁰  =  0 ^n-n  =  0ⁿ  / 0ⁿ   =    0  / 0

But also note that    0  / 0   =  pi        or  0 / 0    =  11   or   0 / 0  = e  ......  etc....

Thus....we  can't actually determine  the value  of  0^n-n  = 0⁰  

Its zero