Whilst I am reading articles on unipotent completion to understand its basic construction, I found something confusing. Let $F$ be a free group of rank 2 whose generating letters are $x$ and $y$ and let $K$ be a field of characteristic 0. Then the unipotent completion of $F$ over $K$ would be the set of group-like elements in the completed Hopf algebra $K[F]^\wedge=\varprojlim K[F]/I^n$ where $I=\ker(K[F]\to K)$ is the kernel of the augmentation map. Note that $K[F]^\wedge$ is isomorphic to the ring of non-commutative formal power series in two variables $K\langle\!\langle X,Y\rangle\!\rangle$, i.e. $K[F]^\wedge\simeq K\langle\!\langle X,Y\rangle\!\rangle$. Now, here is the point I am confusing: why is the natural embedding $F\to K[F]^\wedge\simeq K\langle\!\langle X,Y\rangle\!\rangle$ given explicitly by $x\mapsto e^X$ and $y\mapsto e^Y$? Is this a chosen one?

More precisely, I guess that the natural embedding $F\to K[F]^\wedge$ would be $x\mapsto x$ and $y\mapsto y$ (up to the abuse of notations). So, I guess that the explicit form is given by the isomorphism $K[F]^\wedge\simeq K\langle\!\langle X,Y\rangle\!\rangle$. But I think that the map $x\mapsto 1+X$, $y\mapsto 1+Y$ also gives an isomorphism, is it wrong?