Poincaré Lemma Contractible Hypothesis Poincaré's Lemma is often stated as saying that a closed differential form on a star-shaped domain is exact. More generally, it is true that a closed differential form on a contractible domain is exact.
What I am wondering is if there is an easy example of a closed differential form on a simply connected domain which is not exact.
 A: Let  $U=\mathbb R^n\setminus \lbrace0 \rbrace\subset \mathbb R^n$, a simply connected domain for $n\geq3$,   which we assume from now on.     
The  $(n-1)$ form $\omega \in \Omega^{n-1}(U)$ defined by 
$$\omega (x)=\frac {1}{\mid \mid x\mid \mid^n}   \sum _{i=1} ^n (-1)^{i-1}x_idx^1\wedge...\wedge \widehat{dx^i}              \wedge dx_n$$ for $x\in \mathbb R^n\setminus \lbrace0 \rbrace)$ is closed but not exact. It is thus an example of what you want.
 More precisely, its cohomology class $[\omega ]$ generates the one-dimensional $(n-1)$-th De Rham cohomology vector space of $U$, namely $H^{n-1}_{DR}(U)=\mathbb R\cdot[\omega ]$
NB
a) This form can also be seen as  the pull-back $\omega =r^*(vol)$ of the canonical volume form $vol\in \Omega ^{n-1}(S^{n-1})$ of the unit sphere under the  map $$r:U\to S^{n-1}:x\mapsto \frac {x}{\mid \mid x\mid \mid}$$
b) To be quite explicit, the value of the alternating form $\omega (x)$ on the $(n-1)$-tuple of vectors $v_1,...,v_{n-1}\in T_x(U)=\mathbb R^n$ is $$\omega (x)(v_1,...,v_{n-1})=\frac {1}{\mid \mid x\mid \mid^n}\cdot det[x\mid v_1\mid ...\mid v_{n-1}]$$
where of course $x, v_1, ...,v_{n-1}$ are seen as column vectors of size $n$.
A: The inverse square vector field $\vec{r}/|\vec{r}|^3$ gives rise to its flux 2-form $$\frac{xdydz+ydzdx+zdxdy}{(x^2+y^2+z^2)^{3/2}}$$ in $R^3$ minus the origin, which is simply connected.
This is closed because the field is the gradient of the harmonic function $1/|\vec{r}|$. It is not exact because its integral over the 2-sphere is not zero.
A: Because specific examples have already been given, I will make some general statements.
I will assume (without proof) that an $n$ dimensional manifold $M$ is orientable if and only if there is a nonvanishing $n$ form on $M$.
Let $M$ be a compact, orientable, $n$ dimensional manifold without boundary. Then, it is a simple application of Stokes' theorem to show that the nonvanishing form is closed but not exact. To answer your question, we just need to find a simply connected manifold with these properties, for example, $S^n$ for $n \geq 2$.
For an even more general theorem (which is much harder to prove), if $M$ is a connected $n$ dimensional manifold, $H^n_{de}(M) = 0$ if $M$ is non-orientable or non-compact and $H^n_{de}(M) = \mathbb{R}$ otherwise (if it is both orientable and compact.)
