Value of Finite Continued Fraction equals Numerator Divided by Denominator

Theorem
The value of $\left[{a_1, a_2, a_3, \ldots, a_n}\right]$ is $\dfrac {p_n} {q_n}$.

Proof
We will use a proof by induction on the number $n$ of partial quotients in the continued fraction expansion:

For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\left[{a_1, a_2, a_3, \ldots, a_n}\right] = \dfrac {p_n} {q_n}$

Basis for the Induction
$P(1)$ is the case:
 * $\left[{a_1}\right] = \dfrac {a_1} 1 = \dfrac {p_1} {q_1}$

This holds for any continued fraction expansion:

$P(2)$ is the case:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\left[{a_1, a_2, a_3, \ldots, a_k}\right] = \dfrac {p_k} {q_k}$

Then we need to show:
 * $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k + 1} }\right] = \dfrac {p_{k + 1} } {q_{k + 1} }$

Induction Step
This is our induction step:

Consider the continued fraction expansion:
 * $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k + 1} }\right]$

The numerators are:
 * $p_1, p_2, p_3, \ldots, p_k, p_{k + 1}$

and the denominators are:
 * $q_1, q_2, q_3, \ldots, q_k, q_{k + 1}$

By the second continued fraction identity:
 * $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k + 1}}\right] = \left[{a_1, a_2, a_3, \ldots, a_{k - 1}, a_k'}\right]$

where $a_k' = a_k + \dfrac 1 {a_{k + 1} }$

Consider the.

Take the continued fraction expansion:
 * $\left[{a_1, a_2, a_3, \ldots, a_{k - 1}, a_k'}\right]$

Its numerators are:
 * $p_1, p_2, p_3, \ldots, p_{k-1}$ and $p_k'$

where $p_k' = \left({a_k + \dfrac 1 {a_{k + 1} } }\right) p_{k - 1} + p_{k - 2}$ by definition.

Its denominators are:
 * $q_1, q_2, q_3, \ldots, p_{k - 1}$ and $q_k'$

where $q_k' = \left({a_k + \dfrac 1 {a_{k + 1} } }\right) q_{k - 1} + q_{k - 2}$ by definition.

As it has just $k$ partial quotients, the induction hypothesis tells us that its value is:
 * $\dfrac {p_k'} {q_k'}$

So:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k + 1} }\right] = \dfrac {p_{k + 1} } {q_{k + 1} }$