Proof of $p_{1}p_{2}\dots p_{n} + 1 \geq p_{n+1}$, $\forall n \in \mathbb{N}, n\geq 2$

Question

I'm trying to prove: $$p_{1}p_{2}\dots p_{n} + 1 \geq p_{n+1}$, $\forall n \in \mathbb{N}, n\geq 2$$

I'm trying to do it by induction... Maybe there's a simpler way I'm not seeing?

Answer 1

(1). If $2\leq n\in \mathbb N$ then $n$ is divisible by a prime.

Proof: Let $x$ be the least natural number that is greater than $1$ and divides $n.$ We have $1<x\leq n,$ and $xy=n$ for some $y\in \mathbb N.$

If $x$ is not prime then $x=x'x''$ where $x'$ and $x''$ are natural numbers, each greater than $1$ and less than $x.$ But then $n=x'x''y$ is divisible by $x',$ with $1<x'<x,$ but that contradicts the definition of $x.$

So $x$ must be prime.

(2). $N=1+\prod_{i=1}^np_i$ is divisible by a prime $p$ by (1). And $p\not\in \{p_1,...,p_n\}$ because $ N $ is not divisible by any $p_1,...,p_n$ (because $ N $ is $1$ more than a multiple of each $p_i$ for $i=1,...,n).$

So $p\geq p_{n+1}$.

So $N\geq p\geq p_{n+1}.$

Remark . Euclid used the method of (2) to show, in modern notation and terminology, that if $S$ is a non-empty finite collection of primes then $1+\prod_{p\in S}p$ is divisible by a prime not in $S,$ so $S$ cannot contain all the primes.

Answer 2

$p_1...p_n+1$ is not divided by $p_1,...,p_n$, but it is divided by a prime number $p_i, i\geq n+1$. This implies that $p_1...p_n+1\geq p_i\geq p_{n+1}$.