Show that $(C^{k}([0,1],F),\left \| . \right \|_{C^{k}})$ is banach , where $(F,\left \| . \right \|_{F})$ is banach

Question

Show that $(C^{k}([0,1],F),\left \| . \right \|_{C^{k}})$ is banach , where $(F,\left \| . \right \|_{F})$ is banach and $\left \| f \right \|_{C^{k}}=\sum_{j=0}^{k}\left \| f^{(j)} \right \|_{\infty}$ $\forall (f_{n})_{n}\subset (C^{k}([0,1],F),\left \| . \right \|_{C^{k}})$ Cauchy ie $\forall \epsilon>0,\exists N\in\mathbb{N},\forall n \geqslant N$ and $\forall p \geq N\Rightarrow \left \| f_{n}-f_p\right \|_{C^{k}}\leqslant \epsilon$. for $x$ fixed, $\left \| f_{n}(x)-f_{p}(x) \right \|_{F}\leqslant \left \| f_{n}-f_{p} \right \|_{\infty}\leqslant \left \| f_{n}-f_{p} \right \|_{C^{k}}$.So $(f_{n}(x))$ is cauchy and $(F,\left \| . \right \|_{F})$ is banach, so $(f_{n}(x))_{n}$ we call it f(x). I'm stuck here. Thank you.