Bound on the derivatives of heat kernel. Let the heat kernel $\Phi:\Bbb R^n\times (0,\infty)$ be defined as
$$
\Phi(x,y):=\frac 1{(4\pi t)^{n/2}}e^{-\frac{|x|^2}{4t}}.
$$
From calculation, I have a feeling the the following should be true:

$$
|\nabla^k \Phi(x,t)|\le \frac {C(n,k)}{t^{k/2}} \left( 1+\frac{|x|^2}t \right)^{k/2} \Phi(x,t).
$$

At least for even  $k$ this seems to be the case. Is this estimate well-known? Is it even true?
I'm trying to prove this but induction seems painful to me (or maybe I'm just bad at calculation). If it is true does anyone have a nice proof of it? A hint is also appreciated. 
 A: Step 1: It suffices to prove the assertion in dimension $n=1$: By the very definition of $\Phi$, we have
$$\Phi(x) = \prod_{i=1}^n \left( \frac{1}{\sqrt{4 \pi t}} \exp \left[- \frac{x_i^2}{2t} \right] \right).$$
Assume that we know that the estimate holds in dimension $n=1$. If $\alpha = (\alpha_1,\ldots,\alpha_n) \in \mathbb{N}_0^n$ is a multi-index, then
$$\partial^{\alpha} \Phi(x) = \prod_{i=1}^n \left( \frac{d}{dx_i^{\alpha_i}} \frac{1}{\sqrt{4 \pi t}} \exp \left[- \frac{x_i^2}{2t} \right] \right)$$
and therefore
$$\begin{align*} |\partial^{\alpha} \Phi(x)| &\leq C \prod_{i=1}^n \left( \frac{1}{t^{\alpha_i/2}} \left[1+ \frac{x_i^2}{t} \right]^{\alpha_i/2} \frac{1}{\sqrt{4\pi t}} \exp \left[- \frac{x_i^2}{2t} \right] \right) \\ &\leq C \frac{1}{t^{|\alpha|/2}} \left(1+\frac{|x|^2}{t} \right)^{|\alpha|/2} \Phi(x). \end{align*}$$
Step 2: Prove of the assertion in dimension $n=1$:
Set $\Phi(x) = e^{-x^2}$, $x \in \mathbb{R}$. First we show by induction that for any $k \geq 0$ there exists a polynomial $p_k$ of degree $\leq k$ such that
$$\Psi^{(k)}(x) := \frac{d}{dx^k} \Psi(x) = p_k(x) \Phi(x). \tag{1}$$
For $k=0$ this is obvious. Now assume that $(1)$ holds for some fixed $k \in \mathbb{N}$. Then
$$\begin{align*} \Psi^{(k+1)}(x) &= \frac{d}{dx} \Psi^{(k)}(x) = p_k'(x) \Phi(x) - 2x p_{k}(x) \Phi(x) = \underbrace{(p_k'(x)-2x p_k(x))}_{=:p_{k+1}(x)} \Phi(x), \end{align*}$$
i.e. $(1)$ holds for $k+1$. This finishes the induction and hence the proof of $(1)$.
By $(1)$ there exists a constant $c=c(k)>0$ such that
$$\Psi^{(k)}(x) \leq c(1+|x|^2)^{k/2} \Psi(x) \qquad \text{for all $x \in \mathbb{R}$.} \tag{2}$$
Since
$$\Phi(x) = \frac{1}{\sqrt{4 \pi t}} \Psi \left(\frac{x}{2 \sqrt{t}} \right)$$
it follows from the chain rule that
$$\begin{align*} \Phi^{(k)}(x) &= \frac{1}{\sqrt{4 \pi t}} \frac{1}{(2\sqrt{t})^k} \Psi^{(k)} \left( \frac{x}{2 \sqrt{t}} \right) \end{align*}$$
and so by (2)
$$|\Phi^{(k)}(x)| \leq C' \frac{1}{t^{k/2}} \left(1+ \frac{x^2}{4t} \right)^{k/2} \Phi(x).$$
