Gevrey estimate of derivatives Let $\phi$ be the $C^\infty$ function defined on $\mathbb R_+^*$ by $e^{-t^{-2}}$ and by $0$ on $\mathbb R_-$. 
Question: I think that there exists $\rho>0$ such that
$$
\forall t\in \mathbb R,\forall n\in \mathbb N,\quad\vert{\phi^{(n)}(t)}\vert\le (n!)^{3/2} \rho^{1+n},
\tag{$\ast$}
$$
but I do not have a simple proof. I would like to know if an elementary argument could provide the above estimates.
As an interesting byproduct of this global Gevrey estimate of order $3/2$, there is this nice counter-example by A.N. Tychonov violating Cauchy uniqueness for the heat equation with
$$
u(x,t)=\sum_{n\ge 0}\phi^{(n)}(t)\frac{x^{2n}}{(2n)!},\quad \partial_t u-\partial_x^2 u=0, \quad u_{\vert t\le 0}=0.
$$
Of course to prove convergence in $C^\infty$ of the series defining $u$, some estimates are needed and $(\ast)$ is sufficient.
 A: Faa di Bruno's formula for derivatives of compositions of functions says
$$
(f\circ g)^{(n)}(t)=n!\sum_{k\ge 0}\ \sum_{n_1,\ldots,n_k\ge 1}
\mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\}
\ \frac{f^{(k)}(g(t))\ g^{(n_1)}(t)\cdots g^{(n_k)}(t)}{k!\ n_1!\cdots n_k!}
$$
where $\mathbf{1}\{\cdots\}$ stands for the indicator function of the condition within braces.
For $f(x)=e^{-x}$ and $g(t)=\frac{1}{t^2}$, we of course have $f^{(n)}(x)=(-1)^n e^{-x}$
and $g^{(n)}(t)=(-1)^n (n+1)!\ t^{-(n+2)}$ for the derivatives.
When $n\ge 1$, we thus get
$$
\phi^{(n)}(t)=e^{-t^{-2}}  n!
\sum_{k\ge 1} \sum_{n_1,\ldots,n_k\ge 1}
\mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\}
\frac{(-1)^{n+k}t^{-(n+2k)}}{k!} (n_1+1)\cdots (n_k+1)
$$
Now use the trivial bound $\frac{x^m}{m!}\le e^x$ for $x=t^{-2}$ (followed by taking the square root) and
for $x=\frac{1}{2}t^{-2}$ in order to get
$$
t^{-n}\le \sqrt{n!}\ e^{\frac{1}{2}t^{-2}}
$$
and
$$
t^{-2k}\le 2^k\ k!\ e^{\frac{1}{2}t^{-2}}\ .
$$
This results in
$$
|\phi^{(n)}(t)|\le
n!^{\frac{3}{2}}\times \sum_{k\ge 1}\ \sum_{n_1,\ldots,n_k\ge 1}
\mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\}
\ 2^k\ (n_1+1)\cdots (n_k+1)\ .
$$
By the arithmetico-geometric inequality and the elementary bound $1+x\le e^x$, we have
$$
(n_1+1)\cdots (n_k+1)\le \left(\frac{n+k}{k}\right)^k\le e^n\ .
$$
Therefore,
$$
|\phi^{(n)}(t)|\le
n!^{\frac{3}{2}}\ e^n\times \sum_{k\ge 1}\ 2^k\ \sum_{n_1,\ldots,n_k\ge 1}\mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\}
$$
$$
=n!^{\frac{3}{2}}\ e^n\times \sum_{k=1}^{n}\ 2^k\ \left(
\begin{array}{c}
n-1 \\
k-1
\end{array}
\right)
=2 n!^{\frac{3}{2}} e^n 3^{n-1}\ .
$$
So your bound holds with $\rho=3e$.
