Proof

Suppose that \(p_1,~p_2,~\ldots,~p_r\) are all existing primes.
Let \(N=p_1p_2\cdots p_r\).
There must exist a prime number \(p\), which divides the natural number \((N-1)\), which also must divide \(N\), by assumption.
From the divisibility law no. 6 it follows from \(p\mid (N-1)\) and \(p\mid N\) that \(p\mid 1\), which is impossible.
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