Derivation of the Compound Interest Formula Without Using Calculus I am a student but I am only in Algebra 2. I am doing a project on compound interest and I am trying to prove how we arrive at the formula $y=Pe^{rt}$ but I am stuck at the step where you get from $y=P(1+r/n)^{nt}$ to $y=P((1+1/n)^{n/r})^{rt}$. I would really appreciate some help with my algebra. Thanks!
Generally, $y=P(1+r/n)^{nt}$ does not imply $y=P((1+1/n)^{n/r})^{rt}$ (Try $P=1, n=2,  r=2, t=2$. It fails, since $P(1+r/n)^{nt}=1(1+\frac{2}{2})^{2 \cdot 2}=16 \neq P((1+1/n)^{n/r})^{rt}=1((1+\frac{1}{2})^{2/2})^{2\cdot 2}=\frac{81}{16})$.  

It looks like something may have been mistyped here.

However, it is true that $$y=P(1+r/n)^{nt}=P((1+r/n)^{n/r})^{rt}.$$

There is a rule for exponentiation that states $(a^b)^c=a^{bc}$. Assuming that $r$ is nonzero, we can write $$y=P(1+r/n)^{nt}=P(1+r/n)^{\frac{n}{r}\cdot (rt)}=P((1+r/n)^{\frac{n}{r}})^{\cdot rt}.$$
For compound interest if the amount at time $t$ is given by $A_t$:
$$A_1 = A_0 + rA_0 = A_0(1 + r)$$
$$A_2 = A_0 + rA_0 + r(A_0 + rA_0) = A_0(1 + r)^2$$
So in general:
$$A_t = A_0(1 + r)^t$$
Where do you want to go from here?