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I am currently working on ways to prove this and got stuck proving it with induction.

Any tips?

How could i prove that for every finite alphabet Σ, ∀ n ∈ ℕ. |Σⁿ| = |Σ|ⁿ?

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  • $\begingroup$ Is it necessary to use induction, or are other proof methods allowable? $\endgroup$ – Luke Mathieson Feb 2 at 12:29
  • $\begingroup$ others are welcome as well $\endgroup$ – user128226 Feb 2 at 13:53
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One way would be to show an isomorphism between strings and ordered sets, then the result is a direct consequence of properties of set cardinalities (namely that $|A \times B| = |A|\cdot|B|$ for any two sets $A$ and $B$).

Let's assume that you don't want to do that (for a start it's probably more work to do it properly - but very quick to do it in a hand-waving manner).

The first step is that we need to show that for a set of string $S$ and an alphabet $\Sigma$, $|S\circ \Sigma| = |S|\cdot|\Sigma|$, where $\circ$ is the concatenation operator (overloaded for sets as well as strings). This is not hard of course: each string $s \in S$ and each $\sigma \in \Sigma$, the string $s \circ \sigma$ is in $S \circ \Sigma$ and this string is different for every choice of $s$ and $\sigma$.

Next we can inductively show that $|\Sigma^{n}| = |\Sigma|^{n}$.

Base Case:

$\Sigma^{0}$ is the set of strings obtained by concatenating elements of $\Sigma$ $0$ times, i.e., just the empty string $\varepsilon$ (or $\lambda$ if you're that way inclined :D). So $|\Sigma^{0}| = 1$. Of course $\alpha^{0} = 1$ for any $\alpha \in \mathbb{N}$, so $|\Sigma^{0}| = 1 = |\Sigma|^{0}$.

Inductive Hypothesis

Assume that $|\Sigma^{k}| = |\Sigma|^{k}$.

Inductive Step

$\Sigma^{k+1} \ = \Sigma^{k}\circ\Sigma$, therefore $|\Sigma^{k+1}| = |\Sigma^{k}\circ\Sigma|$, which we have established is $|\Sigma^{k}|\cdot|\Sigma|$, by the inductive hypothesis (and power laws) $|\Sigma^{k}|\cdot|\Sigma| = |\Sigma|^{k}\cdot|\Sigma| = |\Sigma|^{k+1}$, which is what we wanted to show.

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