I am trying to understand the proof for the following claim.
If $A,X\in M_n(K)$ such that $AX=XA$, then $d_Aexp(X)=exp(A)X$.
Proof: Consider the polynomial function $$P_k:M_n(K)\to M_n(K)\\ A\mapsto A^k$$If $AX=AX$, then $$(A+tX)^k=\sum_{j=0}^{k}{ k\choose k -j}A^{k-j}(tX)^j=\sum_{j=0}^{k}{ k\choose k -j}t^jA^{k-j}X^j$$ (This is the part I don't get) Therefore, $$d_AP(X)=\frac{d}{dt}\Bigr|_{\substack{t=0}}(A+tX)^k=kA^{k-1}X$$.
Then the differential is obtained by differentiating the terms individually in the exponential series.
I don't understand both the equalities in the last equation. Any ideas?