Consider an annual interest rate $r$ paid once annually (m = 1).

• In $1$ year, £$1$ becomes £$(1+r)$
• In $2$ years, £$1$ becomes £$(1+r)(1+r)$
• In $3$ years, £$1$ becomes £$(1+r)(1+r)(1+r)$
• In $n$ years, £$1$ becomes £$(1+r)^n$

This is known as discrete compounding. If payments are received $m$ times per year (instead of once annually), in one year,

$$\Bigl(1 + \dfrac{r}{m} \Bigr)^m$$
becomes the value of £1 at the end of one year.

Continuously compounded interest rates

Let $m \rightarrow \infty$ in one year.

Given that \begin{align*} x &= e^{\text{log }x}\\
\end{align*}we can write
$$\Bigl(1+\frac{r}{m}\Bigr)^m = e^{\text{log }\bigl(1+\frac{r}{m}\bigr)^m}= e^{m \text{log }\bigl(1+\frac{r}{m}\bigr)}$$ and given that,
$$\text{log }(1 + x) = x \; – \dfrac{x^2}{2}+ \dfrac{x^3}{3}- \dfrac{x^4}{4} + \dots $$

The value of £1,
\begin{align*}\Bigl(1+\frac{r}{m}\Bigr)^m &= e^{m \; (\frac{r}{m} \; – \frac{(r/m)^2}{2}+ \frac{(r/m)^3}{3} – \frac{(r/m)^4}{4} + \dots)}\\
& = e^{r \; – \frac{1}{2}\frac{r^2}{m}+ \frac{1}{3}\frac{r^3}{m^2} – \frac{1}{4} \frac{r^4}{m^3} + \dots}\\
& \approx e^r \text{ as } m \rightarrow \infty \text{ in 1 year }
\end{align*}
In $n$ years, £1 becomes \begin{align*}\Bigl(1+\frac{r}{n \times m}\Bigr)^{n \times m } &= e^{n m \text{log }\bigl(1+\frac{r}{m \times n}\bigr)}\\
& = e^{rn}
\end{align*}
Conversely, how much will become £1 in $n$ years?
\begin{align*}
& = e^{-rn}
\end{align*}
Key Assumption:
Interest rates $r$ are constant.