finding the partial bell polynomial of $e^x$ $$
\left(e^{x+z} - e^x\right) = \sum_{n=1}^\infty \frac{z^n}{n!} \frac{d^n}{dx^n}[e^x]
$$
$$
\left(e^{x+z}-e^x\right)^k = \sum_{n \geq k} Y^{\Delta}_{e^x}(n,k,x)z^n
$$
Where:
$$
Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{e^x}(x)
$$
$$
[z^n]\left(e^{x+z}-e^x\right)^k = e^{kx}[z^n]\left(e^z-1\right)^k = e^{kx} \sum_{j=0}^k {k \choose j} [z^n]e^{zj} (-1)^{k-j}
$$
$$
e^{zj} = \sum_{n=0}^\infty \frac{(zj)^n}{n!}
$$
therefore:
$$
[z^n]e^{zj} = \frac{j^n}{n!}
$$
$$
B_{n,k}^{e^x}(x) = \frac{e^{kx}}{k!} \sum_{j=0}^k {k \choose j} j^n (-1)^{k-j}
$$
Can someone please check my work, im a bit wary about this...

I think your calculation is correct.

Observe that
\begin{align*}
B_{n,k}^{e^x}(0)=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}j^n(-1)^{k-j}=\left\{n \atop k\right\}
\end{align*}
yield the Stirling Numbers of the second kind.

