Exponent Challenge

By law of indices, we have $a = a^1$ , $a^{-1} = \dfrac1a$ , and $a^{m-n} = a^m \div a^n$ . We have

$\Large \dfrac{2015^{2015}}{2015} = 2015^{2015} \times 2015^{-1} =\boxed{ 2015^{2014}}.$

When bases are same under division powers get subtracted

But what is the exact answer? I got 10461721550869375

$\frac{2015^{2015}}{2015}=2015^{2015-1}=2015^{2014}$

We have: 2015^2015÷2015

= 2015 X 2015 X 2015 X........(2015 times)÷2015

Cancelling 2015 from numerator and denominator

= 2015 X 2015 X 2015 X .........(2014) times = 2015^2014

((2015)^2015)/2015=2015^2014

In a fraction, if the numerator and the denominator are the same, then their powers get subtracted. As in this case, the number being 2015, it's powers are subtracted, i.e. 2015-1=2014.

x^(x)/x=x^(x-1) Plug 2015 into the formula to obtain the solution.

as bases are same, so it can be solved as 2015^(2015-1)=2015^2014

2015^2015 = 2015^2014 x 2015 :v then 2015 in numerator goes with 2015 in denominator So the result is 2015^2014 :v :D

2015^2015/2015=2015^(2015-1) = 2015^2014 (Ans)

We see 2015^2015/2015 that the exponent of the numerator is higher than the exponent of the denominator, so all we need to do is subtract the exponent of the numerator(2015) to the exponent of the denominator(1) sol= (2015) - (1) is equal to 2014, copy the base since their base is common. Therefore 2015^2015/2015 = 2015^2014

We would just reduce a power from the numerator since the bases of the numerator and denominator are same

$\frac{2015^{2015}}{2015}=\frac{2015^{2014}\times2014}{2014}=\boxed{2015^{2014}}$

For n= 1

$\frac{1^{1}}{1} = 1$

n=2 $\frac{2^{2}}{2} = 2$

n=3 $\frac{3^{3}}{3} = 3$

so, $\frac{n^{n}}{n} = n-1$