Higher derivatives of the map $I:T \mapsto T^{-1}$, where $T \in \mathcal B(X)$. Let $X$ be a Banach space, $\mathcal B(X;Y)$ denotes the set of bounded linear operators $X\to Y$. Consider the inverting map $I:U\subset\mathcal B(Y;X)\to \mathcal B(X;Y)$ defined by $I(T) = T^{-1}$, where $U$ is the set where this makes sense. It is known, e.g. here, that $I$ is (Frechet) differentiable and
$$
I'(T)[A] = -T^{-1}AT^{-1}, 
$$
here $I'(T)$ is viewed as an element of $\mathcal B(\mathcal B(Y;X);\mathcal B(X;Y))$.

How do we prove that the $k^{\text{th}}$-derivative of $I$ is the $k$-multilinear map
  $$
(A_1,\dots,A_k) \mapsto (-1)^{k} \sum_{\sigma\in S_k} T^{-1}A_{\sigma(1)}T^{-1}\dots T^{-1}A_{\sigma(k)} T^{-1},
$$
  where the sum is over all permutations $\sigma$ of $\{1,\dots,k\}$?

This formula is given in a book by Hormander without a proof (as usual). It looks like a symmetrization of the higher order terms in the Taylor expansion of $I$ (some details are seen in this thread). 
To obtain higher order derivatives, I tried to differentiate $I'$ by writing $I' = -M\circ I$, where $M(T)[A] = TAT$, and repeatedly apply chain rule. However, the higher derivatives of $M$ gets ugly really fast (or that I don't know a clean way to write it down). Is there a nice way to prove this result?
 A: For small enough $t$ we have the power series expansion
$$ f(T + tS) = (T + tS)^{-1} = (T(1 - (-tT^{-1}S)))^{-1} = T^{-1} \sum_{k=0}^{\infty} (-1)^k (T^{-1}S)^k t^k. $$
Using it we see that,
$$ (D^k f)|_{T}(S, \dots, S) = \left( \frac{d}{dt} \right)^k f(T + tS)|_{t=0} = k! (-1)^k (T^{-1} S)^k. $$
Now we know that $D^k f|_{T}(S_1,\dots,S_k)$ is symmetric and uniquely determined from $D^k f|_{T}(S,\dots,S)$ by the polarization identity so we can just guess that
$$ D^k f|_{T}(S_1,\dots,S_k) = (-1)^k \sum_{\sigma \in S_k} T^{-1} S_{\sigma(1)} \cdots T^{-1} S_{\sigma(k)} $$
and since this is symmetric in $S_1,\dots,S_k$ and coincides with our expression when $S_1 = \dots = S_k = S$, our guess must hold.
A: I decided to post an answer to my own question as an addition to an already good answer by levap. My method will be a direct derivation based on the fact that $I(T)[A] = -T^{-1}AT^{-1}$, using induction.
The base case $k=1$ is already covered by the above formula (whose proof can be found here). Now, assume that 
$$
I^{(k)}(T)[A_1,\dots,A_k] = (-1)^{k} \sum_{\sigma\in S_k} T^{-1}A_{\sigma(1)}T^{-1}\dots T^{-1}A_{\sigma(k)} T^{-1},
$$
holds ($S_k$ is the symmetric group of order $k$). I will rewrite it as 

$$
I^{(k)}(T)[A_1,\dots,A_k] = (-1)^{k} \sum_{\sigma\in S_k} (M_{k,\sigma}\circ I)(T)[A_1,\dots,A_k],
$$

where $M_{k,\sigma}(T)$ is the $k$-linear map
$
M_{k,\sigma}(T)[A_1,\dots,A_k]  = T A_{\sigma(1)}T \dots T A_{\sigma(k)} T.
$
After a little bit of calculation, we can see that 
$$\begin{align}
M_{k,\sigma}(T+S)&[A_1,\dots,A_k] - M_{k,\sigma}(T)[A_1,\dots,A_k] \\
= \ \ \ & (S A_{\sigma(1)}TA_{\sigma(2)}T \dots T A_{\sigma(k)} T)  + (T A_{\sigma(1)}S A_{\sigma(2)}T \dots T A_{\sigma(k)} T) + \dots \\
&\ \ \ + (T A_{\sigma(1)}TA_{\sigma(2)} T \dots T A_{\sigma(k)} S) + o(||S||),
\end{align}$$
which implies that the derivative of $M_{k,\sigma}$ is given by
$$\begin{align}
M'_{k,\sigma}(T)[A_1,\dots,A_k,B] &= (B A_{\sigma(1)}TA_{\sigma(2)}T \dots T A_{\sigma(k)} T) + (T A_{\sigma(1)}B A_{\sigma(2)}T \dots T A_{\sigma(k)} T) + \dots \\
&\ \ \ + (T A_{\sigma(1)}TA_{\sigma(2)} T \dots T A_{\sigma(k)} B).
\end{align}$$
By the chain rule (for multilinear maps), we have 
$$\begin{align}
(M_{k,\sigma}\circ I)'(T)[A_1,\dots,A_k,B] &= (M'_{k,\sigma}\circ I)(T)[A_1,\dots,A_k,I'(T)[B]] \\
&= (M'_{k,\sigma})(T^{-1})[A_1,\dots,A_k,-T^{-1}BT^{-1}] \\
&= (-T^{-1}BT^{-1}) A_{\sigma(1)}T^{-1} A_{\sigma(2)}T^{-1} \dots T^{-1} A_{\sigma(k)} T^{-1} + \dots \\
&\ \ \ \ + T^{-1} A_{\sigma(1)}T^{-1} A_{\sigma(2)} T^{-1} \dots T^{-1} A_{\sigma(k)} (-T^{-1}BT^{-1})
\end{align}$$
Lastly, we apply the above formula to the inductive step to get 
$$\begin{align}
I^{(k+1)}(T)[A_1,\dots,A_k,A_{k+1}] &= (-1)^{k} \sum_{\sigma\in S_k} (M_{k,\sigma}\circ I)'(T)[A_1,\dots,A_k,A_{k+1}] \\
&= (-1)^{k} \sum_{\sigma\in S_k} 
(-T^{-1}A_{k+1}T^{-1} A_{\sigma(1)}T^{-1} A_{\sigma(2)}T^{-1} \dots T^{-1} A_{\sigma(k)} T^{-1} - \dots \\
&\quad\quad\quad\quad\quad\quad  - T^{-1} A_{\sigma(1)}T^{-1} A_{\sigma(2)} T^{-1} \dots T^{-1} A_{\sigma(k)}T^{-1}A_{k+1} T^{-1}) \\
&= (-1)^{k+1} \sum_{\rho\in S_{k+1}} (M_{k+1,\rho}\circ I)(T)[A_1,\dots,A_{k+1}]
\end{align}$$
where the last equality can be verified readily that such permutations really go through all $\rho\in S_{k+1}$.  This concludes the proof.
