Given a positive integer $n$, for all positive integers $a_1, a_2, \cdots, a_n$ that satisfy $a_1 = 1$ and $a_{i+1} \leq a_i + 1$, we aim to find the sum $\displaystyle \sum_{i=1}^{n} a_1 a_2 \cdots a_i$.

To solve this problem, we denote $f(m, n)$ as the sum $\sum a_1 a_2 \cdots a_n$ where $a_1 = m$ and $a_{i+1} \leq a_i + 1$.

We observe that if $a_1 = m$, then $a_2$ can take values from $1$ to $m+1$. Therefore, we have:
$$f(m, n+1) = m \cdot \sum a_2 a_3 \cdots a_{n+1} = m \left( f(1, n) + f(2, n) + \cdots + f(m+1, n) \right).$$

We proceed by induction to find a general formula for $f(m, n)$:
$$f(m, n) = \binom{2n-1}{m+2n-2} \cdot (2n-1)!!.$$

Base Case:
When $n = 1$, it is clear that $f(m, 1) = m$, which satisfies the formula.

Inductive Step:
Assume the formula holds for $n$. Then for $n+1$:
$$\begin{align*} f(m, n+1) &= m \left( f(1, n) + f(2, n) + \cdots + f(m+1, n) \right) \\ &= m \cdot (2n-1)!! \left( \binom{2n-1}{2n-1} + \binom{2n-1}{2n} + \cdots + \binom{2n-1}{m+2n-1} \right). \end{align*}$$

Using the Hockey Stick Identity:
$$\binom{2n-1}{2n-1} + \binom{2n-1}{2n} + \cdots + \binom{2n-1}{m+2n-1} = \binom{2n}{m+2n}.$$

Thus:
$$\begin{align*} f(m, n+1) &= m \cdot (2n-1)!! \cdot \binom{2n}{m+2n} \\ &= m \cdot (2n-1)!! \cdot \frac{(m+2n)!}{m!(2n)!} \\ &= (2n+1)!! \cdot \frac{2n+1}{2n+m}. \end{align*}$$

This confirms that the formula holds for $n+1$.

Therefore, for $m = 1$, we have:
$$f(1, n) = (2n-1)!!.$$

Hence, the desired sum is:
$$\sum_{i=1}^{n} a_1 a_2 \cdots a_i = (2n-1)!!.$$

The answer is: \boxed{(2n-1)!!}.