Prove that $\prod_{i=1}^{n}i^{n+1-i}=\prod_{i=1}^{n}(n+1-i)^i$ by induction.

Question

I'm trying to prove that $$\prod_{i=1}^{n}i^{n+1-i}=\prod_{i=1}^{n}(n+1-i)^i$$ by induction. I've proved it already a different way, but I'm interested to see what a proof by induction would look like for this problem.

Answer 1

Induction goes as follows:

• For $n=2$, obviously the equality is true (the base case)

• Assume that for $n\leftarrow n$, the equality is still true (the hypothesis)

• Now we will prove for $n \leftarrow n+1$. So, we need to prove the following statement: $$\prod_{i=1}^{n+1} i^{n+2-i} = \prod_{i=1}^{n+1} (n+2-i)^i$$ The product on the LHS can be written as: $$\prod_{i=1}^{n+1} i^{n+2-i} =\prod_{i=1}^{n+1} i \cdot \prod_{i=1}^{n+1} i^{n+1-i}=(n+1)!\prod_{i=1}^{n+1} i^{n+1-i}$$ So now, we need to prove $$(n+1)!\underbrace{\prod_{i=1}^{n+1} i^{n+1-i}}_{(*)} = \prod_{i=1}^{n+1} (n+2-i)^i$$ Applying the base case for $(*)$, the equality takes the form $$(n+1)!\underbrace{\prod_{i=1}^{n}(n+1-i)^i}_{(**)} = \prod_{i=1}^{n+1} (n+2-i)^i$$ Now change the bounds of $(**)$ as follows: $$(n+1)!\prod_{i=2}^{n+1}(n+2-i)^{i-1} = (n+1) \prod_{i=2}^{n+1} (n+2-i)^i$$ This comes down to the equality $$n! = \cfrac{\prod_{i=2}^{n+1}(n+2-i)^i}{\prod_{i=2}^{n+1}(n+2-i)^{i-1}}=\prod_{i=2}^{n+1}(n+2-i)$$ which is obviously true.