Suppose I borrow \(B_0\) at an annual interest rate of \(r\).
I repay the loan by making constant payments of \(P\) at the end of each year for \(N\) years.
So, after \(n\in\{0,1,\ldots,N\}\) years, the remaining balance on my loan equals
$$B_n\equiv\frac{P}{r}\left(1-\frac{1}{(1+r)^{N-n}}\right).$$
Although I make the same payment \(P\) each year, the share
$$\begin{align*} S_n &\equiv \frac{B_{n-1}-B_n}{P} \\ &= \frac{1}{r}\left[\left(1-\frac{1}{(1+r)^{N-(n-1)}}\right)-\left(1-\frac{1}{(1+r)^{N-n}}\right)\right] \\ &= \frac{1}{(1+r)^{N-n+1}} \end{align*}$$
of my payment that goes toward the principal rises each year.
Indeed it rises at rate
$$\begin{align*} \frac{S_{n+1}-S_n}{S_n} &= \frac{(1+r)^{-(N-(n+1)+1)}-(1+r)^{-(N-n+1)}}{(1+r)^{-(N-n+1)}} \\ &= r. \end{align*}$$
Intuitively, each payment lowers the remaining balance on which interest accrues.
So the interest added to my loan falls each year, raising the share of my constant payments that go toward the principal.
Thanks to Jeremy Bulow for inspiring this post.