Python 2 - 56 bytes

The first actual answer, if I'm correct. Uses Newton's method.

n=x=input();exec"x-=(x**n-n)/(1.*n*x**-~n);"*999;print x

#Python 2 - 56 bytes

The first actual answer, if I'm correct. Uses Newton's method.

n=x=input();exec"x-=(x**n-n)/(1.*n*x**-~n);"*999;print x

The first actual answer, if I'm correct. Uses Newton's method.

n=x=input();exec"x-=(x**n-n)/(1.*n*x**-~n);"*999;print x

added 3 characters in body

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edited Feb 24, 2016 at 0:03

#Python 32 - 5356 bytes

The first actual answer, if I'm correct. Uses Newton's method.

lambda n,x=1:execn=x=input("x);exec"x-=(x**n-n)/(n*x**1.*n*x**-~n);"*999)and;"*999;print x

#Python 3 - 53 bytes

The first actual answer, if I'm correct. Uses Newton's method.

lambda n,x=1:exec("x-=(x**n-n)/(n*x**-~n);"*999)and x

#Python 2 - 56 bytes

The first actual answer, if I'm correct. Uses Newton's method.

n=x=input();exec"x-=(x**n-n)/(1.*n*x**-~n);"*999;print x

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created Feb 23, 2016 at 23:49

#Python 3 - 53 bytes

The first actual answer, if I'm correct. Uses Newton's method.

lambda n,x=1:exec("x-=(x**n-n)/(n*x**-~n);"*999)and x

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