Let $S(\mathbb{R}^{n})$ be the Schwartz space on $\mathbb{R}^{n}$. For $f\in S(\mathbb{R}^{n})$ and $\alpha=(\alpha_1,\cdots,\alpha_n )\in\mathbb{N}$ and $|\alpha|=\alpha_1+\cdots+\alpha_n$.
Prove by induction that $$D^{\alpha}\hat{f}(x)=(-2\pi i)^{|\alpha|}[\widehat{(\cdot)^{\alpha}f(\cdot)}](x)$$ where $D$ is the partial differential operator given by $$D^{\alpha}f=\frac{\partial^{\alpha}}{\partial x^{\alpha}}f=\left(\frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}\cdots\frac{\partial^{\alpha_2}}{\partial x_2^{\alpha_2}} \frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\right)f$$
Now since $|\hat{f}(x)|=|\int_{\mathbb{R}^n}f(\xi)e^{-2\pi ix\xi}d\xi|\leq \int_{\mathbb{R}^n}|f(\xi)|d\xi$ which is the $L^1$ norm, all derivatives of $\hat{f}$ will be $L^{\infty}$, but I'm not too sure how I should proceed to prove
Hints: You want to induct on the order of the multi-index. For the base case $(|\alpha|=1)$, $D^\alpha=\partial_j$ for some $1\leq j\leq n.$ Check that you get the right result by differentiating under the integral sign.
Now, suppose that it holds for all $|\alpha|=m,$ and consider $D^\beta$, where $|\beta|=m+1.$ Observe that it must be the case that $D^\beta=D^\alpha D^\gamma$, where $|\alpha|=m$ and $|\gamma|=1.$ Apply the base case and the inductive step to conclude.
By the way, it's customary to use $x$ to denote the variable of the original function and $\xi$ to denote the variable of the Fourier-transformed function.