This seems to require a similar answer to an earlier post of mine, so notwithstanding my use of the same hammer for another nail, here goes.
In short, the following just proves that the growth rate of the principal repayment is equal to the periodic interest rate. With that the principal repayment schedule can be calculated quite simply.
The balance of a loan follows this recurrence equation.
p[n + 1] = p[n] (1 + r) - d
where
p[n] is the balance of the loan in month n
r is the monthly interest rate
d is the regular monthly payment
This can be solved like so (using Mathematica in this instance).
RSolve[{p[n + 1] == p[n] (1 + r) - d, p[0] == s}, p[n], n]
where s is the initial loan principal
yielding p[n_] := (d - d (1 + r)^n + r (1 + r)^n s)/r
This notation expresses a formula for the balance in month n, which can be used in a function for the principal repayment pr
, (that is, the regular repayment less the payment of interest on the previous month's balance).
pr[n_] := d - (p[n - 1] r)
Combining these expressions it follows that
$\text{pr}(n)=d (r+1)^{n-1}-r s (r+1)^{n-1}$
So for example, taking a £1000 loan over 3 years with 10% interest per month (rather high, but it's just an example), the monthly repayment d
by standard formula is
s = 1000
r = 0.1
n = 36
d = (r (1+r)^n s)/(-1+(1+r)^n) = 103.34306381837332
Using these figures in a calculation of the principal repayment schedule:
s = 1000
r = 0.1
n = 36
d = 103.34306381837332
pr[1]
3.3430638183733237
pr[36]
93.94823983488523
ListPlot[Array[pr, 36]]

This plots the schedule of principal repayment over the 3 year term.
Consider the growth factors for two sample periods, from month 4 to 5 and from 35 to 36.
pr[5]/pr[4] - 1
0.1
pr[36]/pr[35] - 1
0.1
This is simply the monthly interest rate.
Put algebraically:
$\frac{\text{pr}(n+1)}{\text{pr}(n)}-1=\frac{d (r+1)^n-r s (r+1)^n}{d (r+1)^{n-1}-r s (r+1)^{n-1}}-1=r$
The growth rate of the principal repayment is equal to the periodic interest rate.
So the principal repayment schedule starts at d - s r
and ends at (d - s r)(1 + r)^(n - 1)
.
d - s r
3.3430638183733237
n = 36
(d - s r)(1 + r)^(n - 1)
93.94823983488523
as calculated earlier.
And of course the expression for the principal repayment simplifies as
$\text{pr}(n)=d (r+1)^{n-1}-r s (r+1)^{n-1}=(d-r s) (r+1)^{n-1}$