Sum of Entries in Lesser Diagonal of Pascal's Triangle equal Fibonacci Number

Theorem
The sum of the entries in the $n$th lesser diagonal of Pascal's triangle equals the $n + 1$th Fibonacci number.

Proof
By definition, the entries in the $n$th lesser diagonal of Pascal's triangle are:
 * $\dbinom n 0, \dbinom {n - 1} 1, \dbinom {n - 2} 2, \dbinom {n - 3} 3, \ldots$

and so the statement can be written:
 * $F_{n + 1} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {n - k} k$

The proof proceeds by strong induction.

For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $F_{n + 1} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {n - k} k$

$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $0 \le j \le r$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:

from which it is to be shown that:
 * $F_{r + 2} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {r + 1 - k} k$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{>0}: F_{n + 1} = \displaystyle \sum_{k \mathop \ge 0} \dbinom {n - k} k$