What are the coefficients of an expanded polynomial with given roots I want to find the coefficients $h$ of an $n^{th}$ order polynomial with given roots $a$. The $n^{th}$ order polynomial is given by the geometric series and summation:
$$
\prod_{k=0}^{n-1} x+a_k = \sum_{k=0}^{n-1} h_kx^k
$$
Where all values of $a$ are known.
What I ultimately need to know how to express the $k^{th}$ coefficient $h_k$ as a function of the roots $a_0,a_1,...$. For example, I have found the following by expanding the product by hand and looking for patterns by inspection:
$$
h_{n-1} = 1
$$
$$
h_{n-2} = \sum_{k=0}^{n-1}a_k
$$
$$
h_{n-3} = \sum_{k=1}^{n-1}a_k\sum_{j=0}^{k-1}a_j
$$
$$
h_{0} = \prod_{k=0}^{n-1}a_k
$$
However, I struggle to find the general form for $h_k$

If I understand correctly, I believe you're looking for what the Vieta's formulas gives. Note your $4$ results for $h_0$, $h_{n-1}$, $h_{n-2}$ and $h_{n-3}$ are shown there.
In particular, it states that for the $n$'th degree polynomial
$$P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0} \tag{1}\label{eq1A}$$
with $a_n \neq 0$ and the $r_i$ for $1 \le i \le n$ roots being real or complex numbers, you have the general expression of
$$\sum _{1\leq i_{1}<i_{2}<\cdots <i_{k}\leq n}\left(\prod_{j=1}^{k}r_{i_{j}}\right)=(-1)^{k}{\frac {a_{n-k}}{a_{n}}} \tag{2}\label{eq2A}$$
for $k = 1, 2, \ldots, n$, and with the indices $i_k$ being sorted in increasing order to ensure each product of $k$ roots is used exactly once.
In your case, the $r_i$ are your $-a_i$ and its $a_i$ are your $h_i$, with their $a_n = 1$ in your particular case. Using your symbols, \eqref{eq2A} becomes
$$\begin{equation}\begin{aligned}
& \sum _{1\leq i_{1}<i_{2}<\cdots < i_{k}\leq n}\left(\prod_{j=1}^{k}(-a_{i_{j}})\right) = (-1)^{k}h_{n-k} \\
&h_{n-k} = \sum _{1\leq i_{1}<i_{2}<\cdots < i_{k}\leq n}\left(\prod_{j=1}^{k}a_{i_{j}}\right)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
