I need to show uniform convergence of $f_n(x)=x^{s-1}(1-\frac{x}{n})^n$. Am i right?
No. Using the uniform convergence to exchange a limit and an integral works well on a fixed, bounded interval. However, here, the bound of the integral is increasing, so you are actually working on the whole positive half-line. To see this, we can rewrite the limit as:
$$\lim_{n \to +\infty} \int_0^{+\infty} x^{s-1}\left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x) dx,$$
where $1_{[0, n]}$ is the function whose value is $1$ on $[0, n]$ and $0$ elsewhere. On an unbounded interval, some mass can escape at infinity even if functions get closer and closer uniformly, so the uniform convergence is not enough.
If you know them, my advice would be to use the monotone convergence theorem or the dominated convergence theorem (the later is more convenient here). You will then have to prove that, for all integer $n \ge 1$ and all real $x \ge 0$:
$$0
\le \left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x)
\le e^{-x}. \ \ \ \ (1)$$
[if you choose to use the monotone convergence theorem, you also need to prove that the sequence of functions we are working with is non-decreasing. It is true, but not very fun to show. The inequalities above are easier.]
If you can't use either of these theorems, you can succeed by a clever use of $\varepsilon$'s, especially is you manage to prove the uniform convergence on any subset $[0, M]$ with $M > 0$. But that is a bit cumbersome.
~~~~~
EDIT: So, if you want to use the last method, here are some additional hints. Let $M > 0$. We can assume without loss of generality that $n \ge M$. We will divide the integral in two parts:
$$\int_0^n x^{s-1}\left(1-\frac{x}{n}\right)^n dx = \int_0^M x^{s-1}\left(1-\frac{x}{n}\right)^n dx + \int_M^{+ \infty} x^{s-1} \left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x) dx.$$
We have two pieces. We can control the first piece if we know that the sequence of functions converges uniformly on all compact sets:
$$\lim_{n \to +\infty} \int_0^M x^{s-1}\left(1-\frac{x}{n}\right)^n dx = \int_0^M x^{s-1} e^{-x} dx.$$
We have also a upper bound on the second piece (provided we have proved $(1)$):
$$0 \le \int_M^{+ \infty} x^{s-1} \left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x) dx \le \int_M^{+ \infty} x^{s-1} e^{-x} dx.$$
But the function $x \mapsto x^{s-1} e^{-x}$ is integrable, so that:
$$\lim_{M \to +\infty} \int_M^{+ \infty} x^{s-1} e^{-x} dx = 0.$$
~~~~~
Let's wrap it up. Choose any $\varepsilon > 0$. Then, you can find a $M>0$ such that:
$$\int_M^{+ \infty} x^{s-1} e^{-x} dx < \varepsilon.$$
In addition, if $n$ is large enough, then:
$$\left| \int_0^M x^{s-1}\left(1-\frac{x}{n}\right)^n dx - \int_0^M x^{s-1} e^{-x} dx \right| < \varepsilon,$$
whence:
$$\left| \int_0^n x^{s-1}\left(1-\frac{x}{n}\right)^n dx - \int_0^{+ \infty} x^{s-1} e^{-x} dx \right| < 2\varepsilon.$$
Best Answer
If $f_n(x)\xrightarrow[n\to\infty]{}f(x)$ uniformly, then \begin{align*} \left\lvert\int_a^bf_n(x)\,\mathrm dx-\int_a^bf(x)\,\mathrm dx\right\rvert&\le\int_a^b\lvert f_n(x)-f(x)\rvert\,\mathrm dx\\&\le\int_a^b\|f_n-f\|_\infty\,\mathrm dx\\&=(b-a)\|f_n-f\|_\infty\\&\xrightarrow[n\to\infty]{}0.\end{align*} However, this is obviously assuming that $(a,b)$ is a finite interval, i.e., both $a$ and $b$ are finite.
Here you are integrating over $(-\infty,\infty)$. Nonetheless, you can use monotone convergence, as $(\Phi(\cdot+n))_{n\ge0}$ is non-decreasing (towards the constant function equals to $1$).
You can also use local uniform convergence as follows. On the one hand, $$\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\le\int_{-\infty}^\infty\phi(x)\,\mathrm dx=1,$$ so $$\limsup_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\le1.$$ On the other hand, for each fixed $N>0$, by the uniform convergence $\Phi(x+n)\phi(x)\xrightarrow[n\to\infty]{}\phi(x)$ on $[-N,N]$: $$\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-N}^N\Phi(x+n)\phi(x)\,\mathrm dx\xrightarrow[n\to\infty]{}\int_{-N}^N\phi(x)\,\mathrm dx,$$ which shows that $$\liminf_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-N}^N\phi(x)\,\mathrm dx.$$ Letting now $N\to\infty$ leads to $$\liminf_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-\infty}^\infty\phi(x)\,\mathrm dx=1.$$ Hence we have $$\lim_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx=1.$$