prove e^pi i = -1
If you'll grant me euler's formula, it is easy.
e^(ix)=cos(x)+i*sin(x) (eulers formula)
let x=pi
e^(i pi)= cos(pi)+i*sin(pi)
observe: cos (pi) = -1
sin (pi) = 0
thus e^(i pi) = -1 + i*0 = -1.
 
    
  
  
  prove e^pi i = -1
If you'll grant me euler's formula, it is easy.
e^(ix)=cos(x)+i*sin(x) (eulers formula)
let x=pi
e^(i pi)= cos(pi)+i*sin(pi)
observe: cos (pi) = -1
sin (pi) = 0
thus e^(i pi) = -1 + i*0 = -1.
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  also 99999999999999999999999999999999999999999999999999999999999999999999999999999999999 cubed times 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
=
9999999999999999999999999999999999999999999999999999999999999999999999999999999999699999999999999999999999999999999999999999999999999999999000000000000000000000000003000000000000000000000000000000000000000000000000000000029999999999999999999999999989999999999999999999999999999999999999999999999999999999700000000000000000000000000000000000000000000000000000000000000000000000000000000001
(388 digits)
To solve this, use a free online calculator like: Big Integer Calculator.
Here is a Quanta Magazine article about some impressive problems being solved in number theory, including work by Oxford mathematician and Fields Medal winner James Maynard.
 
    
  
  
  https://projecteuler.net/ has some nice problems. https://projecteuler.net/problem=656 is a nice one.
 
    
  
  
  prove fermat's last theorem
💀
its already proved
https://www.scienzamedia.uniroma2.it/~eal/Wiles-Fermat.pdf
warning: google says its dangerous to visit so;

also 99999999999999999999999999999999999999999999999999999999999999999999999999999999999 cubed times 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999