Let $n(n\geq2)$ be a  natural number and $a_1,a_2,...,a_n$ natural positive real numbers. Determine the least possible value of the expression  $$E_n=\frac{(1+a_1)\cdot(a_1+a_2)\cdot(a_2+a_3)\cdot...\cdot(a_{n-1}+a_n)\cdot(a_n+3^{n+1})} {a_1\cdot a_2\cdot a_3\cdot...\cdot a_n}$$
                                      
To determine the least possible value of the expression 
\[ E_n = \frac{(1 + a_1) \cdot (a_1 + a_2) \cdot (a_2 + a_3) \cdot \ldots \cdot (a_{n-1} + a_n) \cdot (a_n + 3^{n+1})}{a_1 \cdot a_2 \cdot a_3 \cdot \ldots \cdot a_n}, \]
we will use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality).

1. **Applying AM-GM Inequality:**
   For any sequences of positive real numbers \( (x_i)_{i=1}^n \) and \( (y_i)_{i=1}^n \), where \( n \in \mathbb{Z}_+ \), the AM-GM Inequality states:
   \[ \sqrt[n]{\frac{x_1}{x_1 + y_1} \cdot \frac{x_2}{x_2 + y_2} \cdot \ldots \cdot \frac{x_n}{x_n + y_n}} + \sqrt[n]{\frac{y_1}{x_1 + y_1} \cdot \frac{y_2}{x_2 + y_2} \cdot \ldots \cdot \frac{y_n}{x_n + y_n}} \leq \frac{\frac{x_1}{x_1 + y_1} + \frac{x_2}{x_2 + y_2} + \ldots + \frac{x_n}{x_n + y_n}}{n} + \frac{\frac{y_1}{x_1 + y_1} + \frac{y_2}{x_2 + y_2} + \ldots + \frac{y_n}{x_n + y_n}}{n} = 1. \]

2. **Rewriting the AM-GM Inequality:**
   This implies:
   \[ (x_1 + y_1) \cdot (x_2 + y_2) \cdot \ldots \cdot (x_n + y_n) \geq \left( \sqrt[n]{x_1 \cdot x_2 \cdot \ldots \cdot x_n} + \sqrt[n]{y_1 \cdot y_2 \cdot \ldots \cdot y_n} \right)^n, \]
   with equality if and only if:
   \[ \frac{y_1}{x_1} = \frac{y_2}{x_2} = \ldots = \frac{y_n}{x_n}. \]

3. **Applying to the given problem:**
   Let \( x_1 = 1 \), \( x_2 = a_1 \), \( x_3 = a_2 \), ..., \( x_n = a_{n-1} \), and \( x_{n+1} = a_n \).
   Let \( y_1 = a_1 \), \( y_2 = a_2 \), ..., \( y_n = a_n \), and \( y_{n+1} = 3^{n+1} \).

   Then:
   \[ (1 + a_1) \cdot (a_1 + a_2) \cdot (a_2 + a_3) \cdot \ldots \cdot (a_{n-1} + a_n) \cdot (a_n + 3^{n+1}) \geq \left( \sqrt[n+1]{1 \cdot a_1 \cdot a_2 \cdot \ldots \cdot a_n} + \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n \cdot 3^{n+1}} \right)^{n+1}. \]

4. **Simplifying the expression:**
   \[ \sqrt[n+1]{1 \cdot a_1 \cdot a_2 \cdot \ldots \cdot a_n} = \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n}, \]
   and
   \[ \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n \cdot 3^{n+1}} = \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} \cdot \sqrt[n+1]{3^{n+1}} = \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} \cdot 3. \]

   Therefore:
   \[ (1 + a_1) \cdot (a_1 + a_2) \cdot (a_2 + a_3) \cdot \ldots \cdot (a_{n-1} + a_n) \cdot (a_n + 3^{n+1}) \geq \left( \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} + 3 \cdot \sqrt[n+1]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} \right)^{n+1} = 4^{n+1} \cdot a_1 \cdot a_2 \cdot \ldots \cdot a_n. \]

5. **Finding the least possible value:**
   Hence,
   \[ E_n \geq 4^{n+1}, \]
   with equality if and only if:
   \[ \frac{a_1}{1} = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \ldots = \frac{3^{n+1}}{a_n}. \]

   Solving these ratios, we find:
   \[ a_k = 3^k \text{ for } 1 \leq k \leq n. \]

The final answer is \( \boxed{ 4^{n+1} } \).