The equation e(iπ) + 1 = 0 is actually a specific instance of a much more powerful equation (though this is certainly the most interesting instance).  It so happens that:

e(a*i) = cos(a) + i*sin(a)

	This is incredibly easy to show using Taylor series.  (If you don’t know Taylor series, I’m not about to teach them right now, sorry, and there’s no better way to do it.)
	Remember that:
cos(x) = 1 – x2/(2!) + x4/(4!) – x6/(6!) + x8/(8!) – x10/(10!) + x12/(12!) – x14/(14!)…
	and:
sin(x) = x – x3/(3!) + x5/(5!) – x7/(7!) + x9/(9!) – x11/(11!) + x13/(13!)…
	and:
ex = 1 + x + x2/(2!) + x3/(3!) + x4/(4!) + x5/(5!) + x6/(6!) + x7/(7!)…

Therefore:
e(a*i) = 1 + a*i – a2/(2!) – i*a3/(3!) + a4/(4!) + i*a5/(5!) – a6/(6!) – i*a7/(7!) + a8/(8!)…
and since:
i*sin(x) = ix – ix3/(3!) + ix5/(5!) – ix7/(7!) + ix9/(9!) – ix11/(11!) + ix13/(13!)…
therefore:
cos(a) + i*sin(a) = 1 + i*a – a2/(2!) – i*a3/(3!) + a4/(4!) + i*a5/(5!) – a6/(6!) – i*a7/(7!) + a8/(8!)… = e(a*i)

	And there you have it.  The equation at the beginning is no more than the instance where a = π.