The only thing i've been able to do is to prove it for 1. How do i go about prvoing it for k+1?
Assume that it is true for $n-1$, i.e. $I_{n-1} < (n-1)!$, then $I_n < nI_{n-1} < n\cdot (n-1)! = n!$, and it is therefore true for all $n$.
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Assume that it is true for $n-1$, i.e. $I_{n-1} < (n-1)!$, then $I_n < nI_{n-1} < n\cdot (n-1)! = n!$, and it is therefore true for all $n$.