It is sufficient to show that the following proposition $P( n)$ is always true by induction on $n$ : $P( n)$: Any product of distinct $n$ natural numbers greater than $1$ has more than $n( n+1) /2$ distinct positive divisors. It is not difficult to verify that $P( n)$ is true for $n=1,2,3,4,5.$ Suppose that $P(n)$ is true for some $n\geq 5 $ Let $A=a_{1} a_{2} \dotsc a_{n+1}$ where $a_{1} ,a_{2} \dotsc a_{n+1}$ are distinct and greater than $1$. First, suppose that $A$ is a power of prime number $p$. It is clear that $p^{( n+1)( n+2) /2} \mid A$, so we have $\tau ( A) >( n+1)( n+2) /2$. Next, suppose that $A$ has at least two distinct prime divisors. Let $e=\min\{\nu _{p}( A) ;\ p\mid A\}$ Let $q$ be a prime divisor of $A$ such that $\nu _{q}( A) =e$. We consider the following two cases $( 1) ,( 2)$. $( 1)$ the number of distinct prime divisors of $A$ is at least $3$. Let $h$ be a positive integer such that $q\mid a_{h}$. Since $P( n)$ is true, we have $\tau ( A/a_{h}) >n( n+1) /2$. So $\tau ( A) \geq n( n+1) /2+( n( n+1) /2)^{2/3} >( n+1)( n+2) /2$. Thus $P( n+1)$ is true. (2) the number of distinct prime divisors of $A$ is exactly $2$. Let $m=\max\{\nu _{q}( a_{i}) |\ 1\leq i\leq n+1\}$. Suppose that $m\geq 2$. Let $h$ be a positive integer such that $q^{2} \mid a_{h}$. Since $P( n)$ is true, we have $\tau ( A/a_{h}) >n( n+1) /2$. So $\tau ( A) \geq n( n+1) /2+2( n( n+1) /2)^{1/2} >( n+1)( n+2) /2$. Finally, suppose that $m=1$. Let $p$ be the largest prime divisor of $A$. Let $t=\#\{j\in \mathbb{N} |\ q\mid a_{j}\}$. Then we have $\nu _{q}( A) \geq t$ and $\nu _{p}( A) \geq t^{2} -( n+2) t+\left( n^{2} +3n+2\right) /2$. So $\tau ( A) \geq ( t+1)\left( t^{2} -( n+2) t+\left( n^{2} +3n+4\right) /2\right)$. Let $f( x)$ be the real-valued function defined by the following : $f( x) =( x+1)\left( x^{2} -( n+2) x+\left( n^{2} +3n+4\right) /2\right)$. We can verify that $f'( x) >0$. So $\tau ( A) \geq f( 1) =n^{2} +n+2 >( n+1)( n+2) /2$. Thus $P( n+1)$ is true in the case $( 2)$. Therefore, $P( n)$ is true for all positive integers $n$. $\blacksquare $