Why is $\int_{0}^{t} e^{nt} \mathrm{\ dt} = \frac{1}{n} \left(e^{nt} - 1\right)$? [solved; notation is also faulty in the first place]

Question

If $e^{nt}$ can also be written as $\left(e^n\right)^t$ or $\left(e^t\right)^n$, $\int_{0}^{t} e^{nt} \mathrm{\ dt}$ can also be written as $\int_{0}^{t} \left(e^{t}\right)^n \mathrm{\ dt}$ which can also be written as … well, what? I need some more steps of calculation in order to understand why the solution is $\frac{1}{n} \left(e^{nt} - 1\right)$.

Answer 1

$$\int_0^t e^{nz} \mathrm{\ d}z=\left[\frac{1}{n}e^{nz}\right]_0^t=\frac{e^{nt}-1}{n}$$

Answer 2

If $n \not= 0$, you are a fan of the power law, and like complicated answers to simple questions you could compute $$\int_0^t e^{nx} \, dx = \int_0^t (e^x)^{n-1} e^x \, dx = \int_1^{e^t} y^{n-1} \, dy = \left[ \frac{y^n}{n} \right]_1^{e^t} = \frac{e^{nt} - 1}{n}$$ using the substitution $y = e^x$, $dy = e^x \, dx$.