# Finite Moments of Vector in Exponential Family

I am studying some notes on exponential families and there is a section on the computation of moments. The exponential family has the form $$\exp(\sum_{j = 1}^k \phi_j B_j(x) + C(x) - D(\phi))$$

I understand that for $$\lvert s \rvert < \delta$$, $$M_{B(X)}(s) = \mathbb{E}[\exp(s^TB(X))] < \infty$$ but don't see how this implies that $$\mathbb{E}[\lVert B(X) \rVert ^k] < \infty$$. (I assume $$\lVert .\rVert$$ is the Euclidean norm)

I considered setting $$s = \delta_0 \frac{B(X)}{\lVert B(X) \rVert}, \delta_0 < \delta$$, to give

$$\mathbb{E}[\exp(s^TB(X))] = \mathbb{E}[\exp(\delta_0\lVert B(X) \rVert]$$

but I think $$s$$ needs to be a constant vector and so this approach is not allowed.

I have seen the following post (Does any moment generating function implies an existence of moments?) which demonstrates how finiteness of the mgf implies that $$\mathbb{E}[\lvert X \rvert^k] < \infty$$, but cannot see how to apply it to $$\mathbb{E}[\lVert B(X) \rVert ^k] < \infty$$.

I would appreciate it if I could be shown how

$$M_{B(X)}(s) < \infty \implies \mathbb{E}[\lVert B(X) \rVert ^k] < \infty.$$