The upper bound on a Nondeterministic Finite Automata's configurations number

In "Engineering: A Compiler" 2nd ed. by Cooper and Torczon,
in 2.4.1 "Nondeterministic Finite Automata" of Chapter 2,

section "Equivalence of NFAs and DFAs"

discusses the upper bound of a NFA's configurations number.

Here

• NFA stands for Nondeterministic Finite Automata

• DFA stands for Deterministic Finite Automata

NFA is a is a five-tuple $(S, \Sigma, \delta, s_0 , S_A)$, where

• $S$ is the finite set of states in the recognizer, along with an error state $s_e$.

• $\Sigma$ is the finite alphabet used by the FA. Typically, $\Sigma$ is the union

of the edge labels in the transition diagram.

• $\delta(s,c)$ is the FA's transition function. It maps each state $s ∈ S$

and each character $c \in \Sigma$ into some next state. In state $s_i$ with input
character $c$, the FA takes the transition $s_i \xrightarrow{\text{c}}\delta(s_i ,c)$.

• $s_0 \in S$ is the designated start state.

• $S_A$ is the set of accepting states, $S_A \subseteq S$.

A configuration of a NFA is defined as follows


Each time the NFA must make a nondeterministic choice, the NFA clones
itself to pursue each possible transition. Thus, for a given input
character, the NFA is in a specific set of states, taken across all of
its clones.


In this model, the NFA pursues all paths concurrently. At
any point, we call the specific set of states in which the NFA is
active its configuration.


When the NFA reaches a configuration in
which it has exhausted the input and one or
more of the clones has reached an accepting state, the NFA accepts the
string.

Given the definitions above, the books states


Consider the state of an NFA when it has reached some point in the
input string. [...] the NFA has some
finite set of operating clones.


The number of these configurations can be bounded;


for each state, the configuration either includes one or
more clones in that state or it does not. Thus, an NFA with $n$ states
produces at most $\lvert\Sigma\rvert^n

Remember that:

We can view a nondeterministic computation as a directed acyclic graph of configurations indexed by time.

I have a guess that in your eyes:

$2^n$ for both, one directs itself and one directs others and at most every has the $\Sigma$ chance to go next state, maybe to itself, maybe the other state, so from $1$ to $n$ state, an NFA produces at most $\mid\Sigma\mid^n$ configurations.

Here are more details: http://www.cse.msu.edu/~torng/360Book/NFA/

StackExchange Computer Science Q#80388, answer score: 2