Playing with i pi

We learn in math that e to the i pi equals -1.  [e in the base of the natural logarithim, i is the square root of minus 1, and pi is the ratio of the area of a circle to the radius squared.]  Squaring both sides gives e to the i pi squared equals 1.  Now e to the i pi squared equals e to the 2 i pi and 1 equals e to the 0.  Have I proven, falsely, that 2 i pi equals 0 so that i equals 0?  Actually not.  What I have proven (again falsely) is that 2 pi equals 0.  It's like saying that since sine of 2 pi equals sine of 0, then 2 pi equals 0.  And if you think in terms of angles, the angle 2 pi is the angle 0.  This is because sine of x is a cyclic function of x. Well, it turns out that e to the i x is also a cyclic function of x for e to the i x equals cos x plus i sin x.  In other words, it is a trigonomic function.