Let me write out my comment so that this question gets an answer. The dominated convergence theorem says the following.
If a sequence of measurable functions $\displaystyle \{f_n(x)\}_{n=1}^{\infty}$, on the measure space $(\Omega, \mathcal{F}, \mu)$, converge point-wise to a function $f(x)$ and every $f_n(x)$ is dominated by some integrable function $g(x)$ i.e. $\lvert f_n(x) \rvert \leq g(x)$, $\forall n$, $\forall x \in \Omega$ and $\displaystyle \int_{\Omega} g d\mu < \infty $, then $\displaystyle \int_{\Omega} f d \mu$ exists and $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n d \mu = \int_{\Omega} f d \mu$$
Note that it is important that the dominating function $g$ is integrable as Henry T. Horton points out in the comments. Also, the condition $\lvert f_n \rvert \leq g$ everywhere, can be relaxed to $\lvert f_n \rvert \leq g$ $\mu$-almost everywhere.
Also, we need $s > -1$. Otherwise the integrals given in the question diverge.
Now lets apply this to the problem at hand. We want to evaluate $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx$.
Define $$f_n(x) = \begin{cases} \left( 1 - \frac{x}n\right)^n x^s & \text{ if }x \in [0, n]\\ 0 & \text{ otherwise}\end{cases}$$
Hence, $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx$.
Now note that $f_n(x)$ converges point-wise to $e^{-x} x^s$ on $[0, \infty)$. This is so since $$\displaystyle \lim_{n \rightarrow \infty} \left(1 - \frac{x}{n} \right)^n = \exp(-x).$$
More importantly, $f_n(x)$ is dominated by $e^{-x} x^s$ on $[0,\infty)$ i.e. $\lvert f_n(x) \rvert \leq e^{-x} x^s$. This follows from the fact that $$1 - t \leq \exp(-t)$$ whenever $0 \leq t \leq 1$. Hence, we get that $$\left(1 - \frac{x}{n} \right) \leq \exp \left(-\frac{x}{n} \right)$$ which in-turn gives us $$\left(1 - \frac{x}{n} \right)^n \leq \exp \left(-x \right).$$
Hence, $\displaystyle f_n(x) \leq \exp(-x) x^s $. Also, $\displaystyle \int_0^{\infty} \exp(-x) x^s dx < \infty$ for all $s > -1$. Hence, in this case, the dominating function is the same as the limit function.
Putting all these things together, we get the desired result.
\begin{align}
\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx & = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx & \text{(From the definition of $f_n(x)$)}\\
& = \int_0^{\infty} \lim_{n \rightarrow \infty} f_n(x) dx & \text{(Since $f_n(x)$ is dominated by $f(x)$)}\\
& = \int_0^{\infty} f(x) dx & \text{(Since $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = f(x)$)}\\
& = \int_0^{\infty} \exp(-x) x^s dx
\end{align}
Consider the sequence of functions on $(0,1)$
$$f_n(x) = \begin{cases}
n & \text{ if } x \in (0,1/n)\\
0 & \text{ otherwise}
\end{cases}$$
We have $\lim_{n \to \infty} f_n(x) = 0 = f(x)$. However,
$$\lim_{n \to \infty} \int_0^1f_n(x)dx = 1 \neq 0 = \int_0^1 f(x) dx$$
The key in the dominated convergence theorem is that sequence of functions $f_n(x)$ must be dominated by a function $g(x)$, which is also integrable.
Best Answer
As Moya stated in the comment section, your dominating function is correct. Indeed it can be shown (and it is in fact well known) that $$\Big(1 - \frac xn\Big)^{n} \uparrow e^{-x}.$$
For the other term notice that the argument in the $\log$ is allowed to range from in $[1,3]$ and $\log x$ in this interval is a positive increasing function, so that you can estimate it with $\log 3$ (you have $3$ multiplying the exponential which is fine, the estimate doesn't have to be sharp.)
Notice that you missed a minus sign when evaluating the integral, the result indeed should be $\log 3$
To answer your second question, notice that for $x \in (0,1)$ we can write $\frac{1}{1 - x}$ as a geometric series, indeed
\begin{align} - \int_0^1\frac{x^p}{1 - x}\log x\,dx = &\ - \int_0^1\sum_{k = 0}^{\infty}x^{k + p}\log x\,dx \\ = &\ \sum_{k = 0}^{\infty} \int_0^1-x^{k + p}\log x\,dx \tag 1\\ = &\ \sum_{k = 0}^{\infty}\frac{1}{(k + p + 1)^2} \\ = &\ \sum_{k = 1}^{\infty}\frac{1}{(k + p)^2} \end{align}
Notice that the crucial step here is $(1)$: to move the sum out of the integral you need to apply the Monotone convergence theorem to the partial sums. This can be done since $-x^{k + p}\log x$ is positive making the partial sums a nonnegative increasing sequence.