Wick power of standard normal random variable equals Hermite polynomial. 
Let $\xi$ be a standard normal random variable, then it's clear
  that :
$$:\xi^n:=h_n(\xi)$$ where $h_n$ is the $n$-th Hermite polynomial.

I understand that $:\xi^n:$ is the orthogonal projection of $\xi^n$ in the $n$-th homogeneous chaos, and that the latter is spanned by the $n$-th hermite polynomial $h_n(\xi)$. Still, I am not able to see why the equality must hold, (does this hold up to a multiplicative constant?)
I mean, does this follow immediately from the definitions?  Because saying that an element belongs to the span of some vector, is not the same as saying that it equals this vector.
Thanks in advance!
 A: Since the Wick product $\mathpunct: \xi_1 \dots \xi_n \mathpunct:$ coincides on any Gaussian Hilbert space containing $\xi_1, \dots, \xi_n$, it is enough to consider the Wick power $\mathpunct: \xi^n \mathpunct:$ on the Gaussian Hilbert space given by the span of $\xi$.
In particular, $H^{\mathpunct: n \mathpunct:}$ is one-dimensional and is spanned by $h_n(\xi)$. Therefore
$$\mathpunct{:}\xi^n \mathpunct{:} = \pi_n(\xi^n) = \lambda h_n(\xi)$$
a.s. for some $\lambda \in \mathbb{R}$.
To identify $\lambda$, notice that by standard properties of orthogonal projections
$$\langle \mathpunct: \xi^n \mathpunct:, h_n(\xi) \rangle = \langle \xi^n, h_n(\xi) \rangle$$
Now since $\xi^n \in \bigoplus_{j=0}^n H^{\mathpunct: j \mathpunct:}$, we have the decomposition $\xi^n = \sum_{j=0}^n \langle \xi^n, h_j(\xi) \rangle h_j(\xi)$. Since the $n$-th order coefficients must match, we see that $\langle \xi^n, h_n(\xi) \rangle = 1$. Therefore
$$\lambda = \langle \mathpunct: \xi^n \mathpunct:, h_n(\xi) \rangle = \langle \xi^n, h_n(\xi) \rangle = 1$$
