# Derivation of formula for continues annuities

I'm currently working through a book on the Theory of Interest. Currently, I'm looking at continuous annuities, or annuities that are paid continuously. The book gives the following formula, the derivation of which makes perfect sense to me. Little delta is the force of interest and nu is the discount factor.

$$a_n = \int_0^n\nu^t$$

$$= \frac{1-\nu^n}{\delta}$$

This I follow so far. The author then states that the formula can be expressed entirely in terms of the force of interest, and gives the following formula but does not show its derivation.

$$a_n= \frac{1-e^{-n\delta}}{\delta}$$

How is this arrived at? I don't understand how the discount factor is equivalent to e to the "-n delta."

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Remember that $\ln(1+i) = \delta$ so that $1+i = e^{\delta}$. Thus, overall, we have
$$v^n = (1+i)^{-n} = \left(e^{\delta}\right)^{-n} = e^{-n\delta}.$$