Is it true that $\left\lvert e^z-\sum\limits_{0\le k\le 10T}\frac{z^k}{k!}\right\rvert\le e^{-9T}$, when $z\in\mathbb C, T\in \mathbb R$ and $\lvert z\rvert\le T$ ?
$\left\lvert\sum\limits_{k>10T}\frac{z^k}{k!}\right\rvert\le \left\lvert\sum\limits_{k>10T}\frac{T^k}{k!}\right\rvert \le e^{-9T}$
$(10T)!\approx\sqrt{2\pi\cdot10T}\left(\frac{10T}{e}\right)^{10T}$, with $\left(\frac{10T}{e}\right)^{10T}\ge(eT)^{10T}$
but how accurate is the Stirling's approximation ? and is it also valid for the complex case ?
The $\approx$ in the Stirling formula $$n!\approx\sqrt{2\pi n}(n/e)^n$$ means $$\lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}(n/e)^n} = 1,$$ and in fact you have $$\sqrt{2\pi}\sqrt{n}(n/e)^n\le n!\le e\sqrt{n}(n/e)^n$$ for al $n$ (see Wikipedia).
About the complex case, you can see Variations on the Stirling's formula for $Γ(z)$, but is totally irrelevant here because $10T\in\Bbb R$.