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}
You can't, it isn't true.
Take your favorite example of a sequence converging pointwise but not in $L^1$. For instance, let's take $X = [0,1]$ with Lebesgue measure; then the sequence $n 1_{[0,1/n]}$ will do. We want this to be $s_n$. Working backwards, we let $f_n = n 1_{[0,1/n]} - (n-1) 1_{[0, 1/(n-1)]}$ (with $f_1 = 1$). Then we have $\sum_{k=1}^\infty f_n(x) = 0$ almost everywhere, and $f=0$ is certainly dominated by a summable (i.e. integrable) function, with $\int f = 0$. On the other hand, $\int f_1 = 1$ and $\int f_k = 0$ for all $k \ge 2$, so $\sum_{k=1}^\infty \int f_k = 1$.
Generally speaking, having control on the limiting function $f$ is never enough in these situations (and the statement "$f$ is dominated by a summable function" is suspicious because it's the same as just saying "$f$ is summable"). You have to have some sort of uniform control over all the approximating functions $f_n$.
A typical condition under which this does hold is to have either $\sum_{k=1}^\infty \int |f_n| < \infty$, or $\int \sum_{k=1}^\infty |f_n| < \infty$ (you can use monotone convergence or Tonelli's theorem to see these are equivalent). In this case you can use either dominated convergence (with dominating function $\sum_{k=1}^\infty |f_n|$) or Fubini's theorem to conclude $\sum_{k=1}^\infty \int f_n = \int \sum_{k=1}^\infty f_n$.
Best Answer
Take $g_n = |g_n| = n\Bbb 1_{\left(\frac{1}{n+1},\frac{1}{n}\right)}$. Since the supports of these functions intersect no where, with $f_n = |f_n|= \sum_1^k g_k$, the best dominating function is the point wise limit
$$f = \sum_{n=1}^∞ n \Bbb 1_{\left(\frac{1}{n+1},\frac{1}{n}\right)} $$ which is not (lebesgue) integrable, $$∫_ℝ f = ∫_0^1 f = \sum_{n=1}^{∞}\int_{\frac{1}{n+1}}^\frac{1}{n}n = ∞$$
A more easily defined family that has the same behavior is $f_n = \frac{\Bbb 1_{|x|>1/n}}{|x|}$.