5. Arrange $\frac{n(n+1)}{2}$ distinct real numbers into a triangular number table, with the $k(k=1,2, \cdots, n)$-th row from top to bottom containing $k$ numbers. Let the maximum number in the $k$-th row be $M_{k}$, then the probability of $M_{1}<M_{2}<\cdots<M_{n}$ is $\qquad$.
First, place $n$ with any other $n-1$ numbers in the $n$-th row, the corresponding probability is $p_{1}=\frac{n}{\frac{n(n+1)}{2}}=\frac{2}{n+1}$; Second, place the largest number among the remaining $\frac{n(n-1)}{2}$ numbers with any other $n-2$ numbers in the $(n-1)$-th row, the corresponding probability is $p_{2}=\frac{n-1}{\frac{n(n-1)}{2}}=\frac{2}{n}$;
Third, place the largest number among the remaining $\frac{(n-1)(n-2)}{2}$ numbers with any other $n-3$ numbers in the $(n-2)$-th row, the corresponding probability is $p_{3}=\frac{n-2}{\frac{(n-1)(n-2)}{2}}=\frac{2}{n-1}$;
And so on, in each step, place the largest number among the remaining numbers with any other $k-1$ numbers in the $k$-th row, until the last number is placed in the first row. Therefore, the probability of event $A=$ " $M_{1}<M_{2}<\cdots<M_{n}$ " is
$$
P(A)=p_{1} p_{2} \cdots p_{n}=\frac{2}{n+1} \cdot \frac{2}{n} \cdots 1=\frac{2^{n}}{(n+1)!} .
$$