We can write, since $f\in\mathcal S(\mathbb R^n)$
\begin{align*}
\lVert x^{\alpha}\partial^{\beta}f\rVert_1&\leqslant
\int_{\mathbb R^n}|x|^{\alpha}|\partial^{\beta}f(x)|dx\\\
&=\int_{\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|\frac 1{(1+|x|)^{n+1}}dx\\\
&\leqslant C'\sup_{x\in\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|
\int_{\mathbb R^n}\frac{dx}{(1+|x|)^{n+1}}\\\
&=C'\sup_{x\in\mathbb R^n}(1+|x|)^{n+1}|x|^{\alpha}|\partial^{\beta}f(x)|
s_n\int_0^{+\infty}\frac{r^{n-1}}{(1+r)^{n+1}}dr,
\end{align*}
where $s_n$ is the area of the unit sphere in $\mathbb R^n$. The last integral
is convergent, and we get the expected result putting $C:=C's_n\int_0^{+\infty}\frac{r^{n-1}}{(1+r)^{n+1}}dr$.
For the second fact, note that $\partial^{\beta}\widehat f(x)=\int_{\mathbb R^n}
i^{\beta}t^{\beta}e^{it\cdot x}f(t)dt$, hence for $x\in\mathbb R^n$:
\begin{align*}
(1+|x|)^N|\partial^{\beta}\widehat f(x)|&=
(1+|x|)^N\left|\int_{\mathbb R^n}e^{it\cdot x}t^{\beta}f(t)dt\right|\\\
&=\sum_{k=0}^N\binom Nk|x|^k\left|\int_{\mathbb R^n}e^{it\cdot x}
t^{\beta}f(t)dt\right|\\\
&=\sum_{k=0}^N\binom Nk\sum_{|\gamma |=k}\left|\int_{\mathbb R^n}
x^{\gamma}e^{it\cdot x}
t^{\beta}f(t)dt\right|\\\
&=\sum_{|\gamma|\leqslant N}\binom Nk\left|\int_{\mathbb R^n}
x^{\gamma}e^{it\cdot x}t^{\beta}f(t)dt\right|.
\end{align*}
Let $\displaystyle I_{\gamma}(x):=\int_{\mathbb R^n}
x^{\gamma}e^{it\cdot x}t^{\beta}f(t)dt$. Integrating by parts and using
Leibniz formula, we have
\begin{align*}
|I_{\gamma}(x)|&=\left|\int_{\mathbb R^n}e^{it\cdot x}\sum_{\alpha\leqslant
\gamma}\binom{\gamma}{\alpha}\partial^{\alpha}f(t)t^{\beta-\alpha}\frac{\beta !}{(\beta-\alpha)!}dt\right|\\\
&\leqslant \beta !\sum_{\alpha\leq \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}\int_{\mathbb R^n}\left|\partial^{\alpha}f(t)t^{\beta-\alpha}\right|dt,
\end{align*}
and using the first point
\begin{align*}
|I_{\gamma}(x)|&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n}
(1+|x|)^{n+1}|x|^{\beta-\alpha}|\partial^{\alpha}f(x)|\\\
&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n}
(1+|x|)^{n+1}(1+|x|)^{\beta-\alpha}|\partial^{\alpha}f(x)|\\\
&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\sup_{x\in\mathbb R^n}
(1+|x|)^{n+1+\beta}|\partial^{\alpha}f(x)|\\\
&\leqslant \beta !\sum_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}\lVert f\rVert_{(n+1+\beta,\alpha)}.
\end{align*}
Putting $A_{\gamma,\beta}=\beta\max_{\alpha\leqslant \gamma}\frac 1{(\beta-\alpha)!}\binom{\gamma}{\alpha}C_{\alpha}$. Then
$|I_{\gamma}(x)|\leqslant A_{\gamma,\beta}\sum_{\alpha\leq\gamma}\lVert f\rVert_{(n+1+\beta,\alpha)}$. Now, put $\displaystyle B_{N,\beta}:=\max_{|\gamma|\leqslant
N}A_{\gamma,\beta}\binom N{|\gamma|}$. We get
\begin{align*}
\lVert \widehat f\rVert_{(N,\beta)}&\leqslant B_{N,\beta}\sum_{|\gamma|\leqslant N}\:
\sum_{\alpha\leq \gamma} \lVert f\rVert_{(n+1+\beta,\alpha)}\\\
&\leqslant B_{N,\beta}\sum_{|\gamma '|\leqslant N} D(\gamma')\lVert f\rVert_{(n+1+\beta,\gamma')},
\end{align*}
where $D(\gamma')$ denote the number of times on which $\gamma'$ is
obtained in the double sum. Finally, we get
$$\lVert \widehat f\rVert_{(N,\beta)}\leqslant C_{N,\beta}\sum_{|\gamma |\leqslant N} \lVert f\rVert_{(n+1+\beta,\gamma)}$$
putting $\displaystyle C_{N,\beta}:=B_{N,\beta}\max_{|\gamma'|\leqslant N}D(\gamma')$.
Best Answer
As Paul Garrett pointed out it is enough to consider $\alpha = \underbrace{(0,\dots,0,1,0,\dots,0)}_{j}$ only, since any other multi-indices follow by induction. Integration by parts gives us $$\begin{align} \widehat{\partial_j f}(\xi) &= \int_{\mathbb{R}^n}(\partial_jf)(x)e^{-2\pi i \langle x,\xi\rangle}dx \\ &= \int_{\mathbb{R}^{n - 1}}\int_{- \infty}^\infty (\partial_jf)(x)e^{-2\pi i \langle x,\xi\rangle}dx_j dx'\\ &= \int_{\mathbb{R}^{n - 1}} f(x)e^{-2\pi i \langle x,\xi\rangle}\big|_{x_j = -\infty}^{x_j = \infty} dx' + 2\pi i\xi_j\int_{\mathbb{R}^n}f(x)e^{-2\pi i \langle x,\xi\rangle}dx\\ &= 2\pi i \xi_j \hat{f}(\xi) \end{align}$$
Since as $f$ is a Schwartz function we have $$|f(x)| \leq C (1 + |x|)^{-1}$$ for some $C > 0$ and thus $$\lim_{x \to \pm\infty}|f(x) e^{-2\pi i x \xi_j}| \leq \lim_{x \to \pm\infty} |f(x)| \leq \lim_{x \to \pm\infty}C (1 + |x|)^{-1} = 0$$