Equivalence of Definitions of Real Exponential Function

Theorem
The following definitions of the exponential function are equivalent.

1 implies 5
This proves that $y$ is a solution.

It remains to be proven that $y$ fulfills the initial condition:

5 implies 1
To solve for $C$, put $\left({x_0, y_0}\right) = \left({0, 1}\right)$:

3 implies 4
From the Binomial Theorem:


 * $\displaystyle \left({1 + \frac x n}\right)^n = 1 + x + \frac{n \left({n-1}\right)x^2}{2! \ n^2} + \frac{n \left({n-1}\right)\left({n-2}\right)x^3}{3! \ n^3} + \cdots$


 * $ = \displaystyle \frac {x^0}{0!} + \frac {x^1}{1!} + \left({\frac {n-1} {n} }\right) \frac {x^2}{2!} + \left(\right) \frac {x^3}{3!} + \cdots$

From Power of Number less than One, this converges to:


 * $\displaystyle \exp x - \frac {x^0}{0!} + \frac {x^1}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots = 0$

as $n \to +\infty$:

Compare Series of Power over Factorial Converges.

3 implies 5
The application of Derivative of Exponential at Zero isn't circular as the referenced proof does not depend on $D_x \exp x = \exp x$.

Also see

 * Equivalence of Definitions of Euler's Number
 * Equivalence of Definitions of Natural Logarithm