Is this an abuse big O notation as a power of a number? Theorem 7.11 in Introduction to the theory of computation 3rth edition says


  Let $t(n)$ be a function where $t(n)&gt;n$. Then every $t(n)$ time
  nondeterministic single-tape Turing machine has an equivalent
  $2^{O(t(n))}$ time deterministic single tape Turing machine.


I can understand that in this proof when we have a nondeterministic TM which every node in it can at most go to $b$  other nodes in its computation tree, then the time complexity of a deterministic Turing machine simulating it would be $O(t(n)b^{t(n)})$. But I'm not sure if I understand the $2^{O(t(n))}$ correctly.

Does it means that if $f(n) \in 2^{O(t(n))}$ then 
$\exists b\in \mathbb{N} \; f(n)\in O(b^{t(n)})?$   

On the other hand, the claim 1.5 in the Computational complexity book says


  For every $f : \{0, 1\}^∗ \rightarrow \{0, 1\}$ and time-constructible
  $T : N\rightarrow N$, if $f$ is computable in time $T(n)$ by a TM $M$
  using alphabet $\Gamma$, then it is computable in time $4 log |\Gamma|T(n)$
  by a TM $M$ using the alphabet $\{0,1,\square,\rhd\}$.


So every nondeterministic TM M running in $t(n)$ time can have an equivalent nondeterministic TM using $\{0,1\}$ alphabet running upper hand in $|\Gamma|^2t(n)$. Thus this equivalent TM's computation tree has at most 2 different branches in each step. Then we can have a deterministic simulator of that TM running in time $O(2^{|\Gamma|^2t(n)})$. 

The intersection of 2 above statement suggests that carelessly we can say $O(2^{t(n)})=2^{O(t(n))}$, but we know $\forall b&gt;2,\; b^{t(n)}\notin O(2^{t(n)})$.

Unfortunately, I didn't find any definition of $2^{O(t(n)}$. I wish to know is there any formal definition for $2^{O(t(n))}$?

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Edit: There is a similar question here. But I'm doubting its answer is the same since the big O notation in the power is not just $O(1)$. 

Suppose we have nondeterministic TM that each node of it goes to 3 other nodes. Then there is a deterministic Turing machine that simulates it in time $t(n)3^{t(n)}$. So we can't say that there is a constant $k$ that $t(n)3^{t(n)} &lt; 2^{kt(n)}.$

Meanwhile, when I think more, this definition holds if we ignore the Sipser proof. I mean if we begin the proof by the statement that every nondeterministic TM has a version $\{0,1\}$ alphabet running in $c.t(n)$ then the answer's definition is correct.

I wonder if there is bad use of big O notation in the Sipser's book or $2^{O(t(n))}$ has a different definition.
This is perfectly standard use of big O notation, despite some purists disliking it. Here is what the passage you quoted means:

Let $t(n)$ be a function where $t(n) &gt; n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{f(n)}$ time deterministic single tape Turing machine, for some function $f(n)$ satisfying $f(n) = O(t(n))$.

This is still somewhat ambiguous, since it’s perhaps not clear what variables the big O depends on. In this case, big O hides a universal constant:

There exists a constant $C&gt;0$ such that the following holds.
Let $t(n)$ be a function where $t(n) &gt; n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{f(n)}$ time deterministic single tape Turing machine, for some function $f(n)$ satisfying $f(n) \leq Ct(n)$ for all $n$.

This allows us to rephrase the statement in yet another way:

There exists a constant $C&gt;0$ such that the following holds.
Let $t(n)$ be a function where $t(n) &gt; n$. Then every $t(n)$ time nondeterministic single-tape Turing machine has an equivalent $2^{Ct(n)}$ time deterministic single tape Turing machine.

(This assumes that by “$f(n)$ time machine” we mean a machine running in time at most $f(n)$.)