We will assume the fact that each term a fixed amount is paid. This amount is #a_j+r_j# for term #j#. We will first show that the installments #a_j# of an annuity loan in the terms #j=1,2,\ldots# form a geometric sequence with scale factor #(1+i)#:

\[\begin{array}{lclcc}
&&Ann = r_j + a_j = r_{j+1} + a_{j+1} &(1)&\\
&&\phantom{xxxxxxx}\color{blue}{\text{characterizing formula annuity }}\\
&&r_j = i \cdot R_{j-1} &(2)&\\
&&\phantom{xxxxxxx}\color{blue}{\text{formula interest payments term } j}\\
&&r_{j+1} = i \cdot R_j = i \cdot (R_{j-1} - a_j) = i\cdot R_{j-1} - i\cdot a_j &(3)&\\
&&\phantom{xxxxxxx}\color{blue}{\text{formula interest payments term }j + 1}&&\\
&&i\cdot R_{j-1} + a_j = i\cdot R_{j-1} - i\cdot a_j + a_{j+1}&&\\
&&\phantom{xxxxxxx}\color{blue}{\text{(2) and (3) entered in second equality (1)}}&&\\
&&a_{j+1}=a_j+i\cdot a_j\\
&&\phantom{xxxxxxx}\color{blue}{\text{the previous equation simplified }}&&\\
&& a_{j+1}=a_j\cdot (1+i) &&\\
&&\phantom{xxxxxxx}\color{blue}{a_j\,\,\text{moved outside brackets }}&&\\
\end{array}\]

This indeed shows #a_j=a_1\cdot (1+i)^{j-1}#. To determine #a_1#, we use the fact that the sum of all installments is equal to the initial borrowed capital #C#:

\[\begin{array}{rcl}C&=&\sum_{j=1}^na_j \\ &=&a_1\cdot\sum_{j=1}^n{(1+i)^{j-1}} \\ &&\phantom{xx}\color{blue}{a_j = a_1\cdot (1+i)^{j-1}}\\ &=&a_1\cdot\dfrac{(1+i)^{n}-1}{1+i-1} \\ &&\phantom{xx}\color{blue}{\text{geometric sequence}}\\ &=&a_1\cdot\dfrac{(1+i)^{n}-1}{i}\end{array}\]

Hence, we find the linear equation #a_1\cdot\dfrac{(1+i)^{n}-1}{i}=C# with unknown #a_1#, leading to

\[a_1 = \dfrac{C\cdot i}{(1+i)^n-1}\]

Now that we know all #a_j#, we calculate the remaining debts #R_j# as follows:

\[\begin{array}{rcl}
R_j &=& R_{j-1}-a_j = R_{j-2}-a_{j-1}-a_j=\cdots\\ &=&R_0-a_1-a_2-\cdots-a_j\\ &=&C-a_1\cdot\left(1+(1+i)+(1+i)^2+\cdots+(1+i)^{j-1}\right)\\ &=&C-a_1\cdot \dfrac{(1+i)^j-1}{i}\\ &&\phantom{xxxxxxx}\color{blue}{\text{geometric sequence }}\\&=&C- \dfrac{C\cdot i}{(1+i)^n-1}\cdot \dfrac{(1+i)^j-1}{i}\\ &&\phantom{xxxxxxx}\color{blue}{\text{formula for }a_1\text{ entered }}\\&=&C- \dfrac{C\cdot \left((1+i)^j-1\right)}{(1+i)^n-1}\\&&\phantom{xxxxxxx}\color{blue}{\text{fractions multiplied }}\\ &=&C\cdot\dfrac{(1+i)^n-1- \left((1+i)^j-1\right)}{(1+i)^n-1}\\&&\phantom{xxxxxxx}\color{blue}{\text{everything on the right hand side brought over one denominator }} \\&=& C\cdot \dfrac{1-(1+i)^{j-n}}{1-(1+i)^{-n}}\\
&&\phantom{xxxxxxx}\color{blue}{\text{fraction simplified}}
\end{array}\]

The interest payments #r_j# are directly related to the remaining debts:

\[
r_j = i\cdot R_{j-1}= i\cdot C\cdot \dfrac{1-(1+i)^{j-1-n}}{1-(1+i)^{-n}}
\]

We indeed conclude that #Ann=a_j+r_j# is constant (meaning: not being dependent on #j#) for the found values of #a_j# and #r_j#:

\[\begin{array}{rcl}
a_j+r_j &=&a_1\cdot (1+i)^{j-1}+i\cdot C\cdot \dfrac{1-(1+i)^{j-1-n}}{1-(1+i)^{-n}}\\&&\phantom{xxxxxxx}\color{blue}{\text{formulas for }a_j\text{ and }r_j\text{ entered }} \\ &=& \dfrac{C\cdot i}{(1+i)^n-1}\cdot (1+i)^{j-1}+i\cdot C\cdot \dfrac{(1+i)^n-(1+i)^{j-1}}{(1+i)^{n}-1}\\ &&\phantom{xxxxxxx}\color{blue}{\text{formula for}a_1\text{ entered }}\\ &=& C\cdot i\cdot \dfrac{ (1+i)^{j-1}+ (1+i)^n-(1+i)^{j-1}}{(1+i)^{n}-1}\\ &&\phantom{xxxxxxx}\color{blue}{\text{like fractions added }}\\ &=& C\cdot i\cdot \dfrac{ (1+i)^n}{(1+i)^{n}-1}\\&&\phantom{xxxxxxx}\color{blue}{\text{terms with }j\text{ cancel out }}\\ &=&\dfrac{C\cdot i}{1-(1+i)^{-n}}\\&&\phantom{xxxxxxx}\color{blue}{\text{numerator and denominator divided by }(1+i)^n} \end{array}\]

The last expression is the formula for#Ann# in the theorem.