Let $A\in B(H)$ be a bounded operator on Hilbert space $H$. Assume that $A = A^{*}$.
Show that for every $t \in \mathbb{R}$ operator defined below is unitary:
$$U(t) = e^{itA} = \sum_{k=0}^{+\infty}\frac{(it)^kA^k}{k!}$$
Using few properties of exponential I have proven before I am able to show that
$$U(t)(U(t))^* = e^{itA}(e^{itA})^* = e^{itA}e^{-itA^*} = e^{itA}e^{-itA} = e^{(it-it)A} = e^{-itA}e^{itA} = (U(t))^*U(t)$$
but I do have some conerns about showing that it behaves like the identity operator.
My attempt:
Let $v \in H$ then
$$(UU^*)v = (e^{0 \cdot A})v = \sum_{k=0}^{+\infty}\frac{A^k0^kv}{k!} = v$$
My question is related to the last equality sing. How do I know that such thing is valid? Since I have undefined symbol $0^0$.
Here , as well as in basic calculus, $\sum a_nx^{n}$ should always be interpreted as $a_0+a_1x+a_0x^{2}+...$. The starting term is $a_0$, not $a_0 x^{0}$, when $x=0$. The starting term in $e^{0}v$ is $Iv=v$.