Prove by induction and recursion that $n!=n(n-1)(n-2)...(3)(2)(1). $

Question

Prove by induction and recursion that $n!=n(n-1)(n-2)...(3)(2)(1). $

We can start with the definition of factorial with recursion:

$$n!= \left\{\begin{align}1\quad \text{for}\quad n=0\\n\cdot(n-1)!\quad \text{for}\quad n>0\end{align}\right.$$

But i can´t see how to use induction to prove it.

Answer 1

We are given that $$n!= \left\{\begin{align}1\quad \text{for}\quad n=0\\n\cdot(n-1)!\quad \text{for}\quad n>0\end{align}\right.$$ and wish to show that $$n! = \prod_{i=1}^ni$$ We start with $n=1 \implies 1! = 1\cdot(0!) = 1 = \prod_{i=1}^1 i$
Now, let's assume $(n-1)! = \prod_{i=1}^{n-1} i$, then $$n! = n\cdot(n-1)! = n\prod_{i=1}^{n-1} i = \prod_{i=1}^{n} i$$

Answer 2

Induction step:

If $$(n-1)!=(n-1).(n-2)\dots3.2.1$$ then by definition $$n!=n.(n-1)!=n.(n-1).(n-2)\dots3.2.1$$