$(1+r)^m$ versus $(1+\frac{r}{n})^{nt}$
I've often in books found two formulations for the compound interest rate formula, which, remember, I see as inclusive of the simple and compounded formulae too. One expression of the compound interest rate - the more messy one - implicitly assumes that $r$ is given to you as an annualised nominal rate, and the other assumes it is simply the nominal rate which applies to the compounding period $m$. The first has more parts to it, due to the initial step of transforming the annualised rate to the compounding period rate. The second doesn't have this step. I prefer this second formula.
It is more basic. We annualise for human purposes. To compare rates across a standardised time horizon. There's nothing inherently mathematically important about the human centric time horizon of a year. For the maths, all that matters is the implicit compounding period, and the number of periods you roll the compounding operation.
So in the formulae at the top of this posting read the first formula as saying the following: Take a dollar and a rate $r$ which has, as all rates do, an implicit compounding basis of some time period. Lay out $m$ of these time measures end to end. Walk that dollar forward and at the end of each of those $m$ time periods, grow the value of your dollar pot by $1+r$. It is a series of $m$ salaries which are paid at the end of some implicit time period, implied in the compounding basis of the nominal rate $r$. The rate doesn't give you a full picture of itself unless accompanied by the compounding basis, a time measure. It is also nominal in the sense that it isn't quite the return you will see on your investment. For example, with a nominal rate of 10% on an annualised compounding basis, run over two years, a dollar becomes 1.21, meaning your holding period return is 21%, not 20% (2 times 10%). Likewise 10% over four years gets you a holding period return of 46.41%, not 40%. Think of 'nominal' as meaning a rate which is used internally in the production of your final holding rate return. It is referenced at $m$ points on this journey. But when someone asks you how much your journey profited you, the answer only emerges after having run the algorithm, made the journey, evaluated the compounding formula.
Simply, if we prefer to see $r$ quoted on an annualised basis, an initial translation needs to happen to make annualised $r$ into the equivalent compounding period rate. Compounding periods are almost always less than a year, so you see $m$ expressed as $m=t \times n$, meaning that the $m$ compounding periods in question run $n$ times a year for $t$ years. So you get to an equivalent compounding period rate by chopping down your annualised rate $r$ to $\frac{r}{n}$
To put this all another way, virtually every time you come to use the compounding formula you'll probably use $(1+\frac{r}{n})^{nt}$ but the mathematics doesn't care about years, and $(1+r)^m$ is the cleaner, clearer formulation. They both do exactly the same job.
