For which $\alpha > 0$ the equality holds: $$ \int_0^1 \sum_{n=0}^\infty x^\alpha e^{-nx} dx= \sum_{n=0}^\infty \int_0^1 x^\alpha e^{-nx} dx $$
We've learned that an interchange can be applied if and only if the $\sum_{n=0}^\infty x^\alpha e^{-nx} $ converges uniformly to some $f$.
So the question translated into "When does the series $\sum_{n=0}^\infty x^\alpha e^{-nx} $ converges uniformly?".
EDIT:
Here's my new realizations:
$$f_n(x) = \sum_{n=0}^\infty x^\alpha e^{-nx}$$
First we'll show a pointwise convergence. Let $x\in (0,1)$:
$$f_n(x) = x^\alpha \sum_{n=0}^\infty e^{-nx} = x^\alpha \sum_{n=0}^\infty \left( \frac{1}{e^x}\right)^n = x^a \frac{e^x}{e^x-1} = f(x)$$
We may also notice that $f_n(x) = f(x)$ where $x=0,1$. Therefore, we have that $f_n(x)$ converges pointwise to $f(x)$ for every $x\in[0,1]$.
I found that $f_n(x)$ has a maximum value when $x=\frac{\alpha}{n}$.
Let us evaluate $f_n(x)$ at that point:
$$f_n\left(x=\frac{\alpha}{n}\right) = \sum_{n=0}^\infty = ... = \left(\frac{\alpha}{n}\right)^\alpha \sum_{n=0}^\infty \frac{1}{n^\alpha}$$
Now, if $\alpha \le 1$ then the series diverges and doesn't equal to $f(x=n/\alpha)$. If $\alpha >1$ then the series converges and equals to $f(x=n/\alpha)$.
So in conclusion, we can interchange if and only if $\alpha > 1$.
Is that all right?
A series of functions is a function. It is absolutely normal that the sum of the series depends on $x$.
The remainder $R_n(x)=\displaystyle\sum_{k\geq n+1}x^\alpha e^{-kx}$ is $R_n(x)=\dfrac{x^\alpha e^{-(n+1)x}}{1-e^{-x}}$. Now you could try to determine the supremum of this guy on $[0,1]$( which will be its maximum since it is a continuous function on a compact interval), or show that it goes to $0$ by other means.
For example if $\alpha>1$, the function $\dfrac{x^{\alpha-1}}{1-e^{-x}}$ is continuous on $[0,1]$, hence bounded, and a quick study of $xe^{-(n+1)x}$ shows that its maximum goes to 0 when $n$ goes to infinity.
I don't have time to do the computations which would allow to compute the maximum in general, but I'm sure that with a litlle patience, you should be able to do it.