I want to show that $$\exp(\sum_{n = 1}^{\infty} \frac{t^n}{n}) = \frac{1}{1 - t},$$ for $t \in (-1, 1)$. Using derivatives and fundamental theorem of calculus, I have a basic idea how to proceed.
But is there a way to do this without using integrals?
I know that $\exp x := \sum_{n = 0}^{\infty} \frac{x^n}{n!}$ and also that this power series and its derivatives converge uniformly on bounded intervals.
We should note by the way that the thing you want to prove is valid for $|t|<1$, not for all $t$.
Say $$f(t)=(1-t)\exp\left(\sum_{n=1}^\infty\frac{t^n}{n}\right).$$You want to show that $f(t)=1$ for $|t|<1$. Since $f(0)=1$ you need only show that $f$ is constant. In other words, $f'(t)=0$ for $|t|<1$.
Now we do need to assume that the derivative of $\sum_{n=1}^\infty t^n/n$ is $\sum_{n=1}^\infty t^{n-1}=1/(1-t)$. The fact that you can differentiate that infinite sum term by term is not quite trivial, but it sounds like it's something that's allowed here. Given that, just go ahead and differentiate $f$ using the product rule and the chain rule; $f'=0$ falls out.