Let $S = \mathbb{C}[x_1,...,x_n]$ be the polynomial ring in $n$ variables, $S_d \subset S$ the subspace of homogeneous polynomials of degree $d$, i.e., the polynomials with the property
\begin{align}
f(tx) = t^d f(x).
\end{align}
Let G be the symmetric group consisting of $n \times n$ permutation matrices. Then $S^G \subset S$ is defined as the subspace of invariant polynomials under G, i.e. the set \begin{align} S^G &= \{f \in S~ |~ f(Ax) = f(x) ~ for ~all ~ A \in G \} \\&= \{ f \in S ~ | ~ f(x_{\sigma(1)},...,x_{\sigma(n)}) = f(x_1,...,x_n) ~for~all~\sigma \in \mathcal{S}_n \} \end{align}
So all the invariant Polynomials are the symmetric Polynomials, which are generated by the $n$ elementary symmetric polynomials \begin{align} \sigma_1 (x) &= \sum_{i}x_i \\ \sigma_2 (x) &= \sum_{i<j} x_i x_j \\ &...... \\ \sigma_n (x) &= x_1 \cdot x_2\cdot... \cdot x_n \end{align}
The Hilbert Series, which I want to find out, is defined as the formal power series \begin{align} P(t) = \sum_{d \geq 0} dim(S^{G} \cap S_d)t^d \in \mathbb{Z}[[t]]. \end{align}
I started calculating the $dim(S^G \cap S_d)$ for each $d$ and I got:
$dim(S^G \cap S_d) = 1 $ for all $d$, since I took the $\sigma_i$ as a basis for each summand $S^G \cap S_d$.
But this must obviously be wrong, because then I get the Hilbert Series \begin{align} \frac{1}{1-t}. \end{align}
In Mukai it says the Hilbert Series for the symmetric Group is
\begin{align} \frac{1}{(1-t)(1-t^2)...(1-t^n)}. \end{align} So I am missing a few factors in my calculation but don't know how to get them. Can anyone show me how to calculate the Hilbert Series in the correct way ?