# Exponential of Zero

## Theorem

Let $\exp x$ be the exponential of $x$.

Then:

$\exp 0 = 1$

## Proof 1

We have that the exponential function is the inverse of the natural logarithm function:

$\ln 1 = 0$

Hence the result.

$\blacksquare$

## Proof 2

Using the definition of the exponential as a limit of a sequence:

 $\displaystyle \exp 0$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 0 n}\right)^n$ $\displaystyle$ $=$ $\displaystyle 1$

$\blacksquare$

## Proof 3

This proof assumes the series definition of $\exp$.

That is, let:

$\displaystyle \exp x = \sum_{k \mathop = 0}^{\infty} \frac {x^k} {k!}$

Then:

 $\displaystyle \exp 0$ $=$ $\displaystyle \sum_{k \mathop = 0}^{\infty} \frac {0^k} {k!}$ $\displaystyle$ $=$ $\displaystyle 1$ Definition of $0^k$

$\blacksquare$

## Proof 4

This proof assumes the Definition of $\exp x$ as the unique continuous extension of $e^x$.

 $\displaystyle \exp 0$ $=$ $\displaystyle e^0$ $\displaystyle$ $=$ $\displaystyle 1$ Definition of $x^0$, where $x \ne 0$

$\blacksquare$

## Proof 5

 $\displaystyle \map \exp {z + \paren {-z} }$ $=$ $\displaystyle \exp z \, \map \exp {-z}$ Exponential of Sum $\displaystyle \leadsto \ \$ $\displaystyle \map \exp {z - z}$ $=$ $\displaystyle \dfrac {\exp z} {\exp z}$ $\displaystyle \leadsto \ \$ $\displaystyle \exp 0$ $=$ $\displaystyle 1$

$\blacksquare$