The sequence ${a_{n}}$ $n\in \mathbb{N}$ is given in a recursive way with $a_{1}=1$,
$a_{n}=\prod_{i=1}^{n-1} a_{i}+1$, for all $n\geq 2$.
Determine the least number $M$, such that $\sum_{n=1}^{m} \frac{1}{a_{n}} <M$ for all $m\in \mathbb{N}$

1. **Define the sequence and initial conditions:**
   The sequence \(\{a_n\}\) is defined recursively with \(a_1 = 1\) and for \(n \geq 2\),
   \[
   a_n = \prod_{i=1}^{n-1} a_i + 1.
   \]

2. **Calculate the first few terms of the sequence:**
   - For \(n = 2\):
     \[
     a_2 = a_1 + 1 = 1 + 1 = 2.
     \]
   - For \(n = 3\):
     \[
     a_3 = a_1 \cdot a_2 + 1 = 1 \cdot 2 + 1 = 3.
     \]
   - For \(n = 4\):
     \[
     a_4 = a_1 \cdot a_2 \cdot a_3 + 1 = 1 \cdot 2 \cdot 3 + 1 = 7.
     \]
   - For \(n = 5\):
     \[
     a_5 = a_1 \cdot a_2 \cdot a_3 \cdot a_4 + 1 = 1 \cdot 2 \cdot 3 \cdot 7 + 1 = 43.
     \]

3. **Define the sum \(b_m\):**
   Let \(b_m = \sum_{n=1}^{m} \frac{1}{a_n}\).

4. **Calculate the first few terms of \(b_m\):**
   - For \(m = 1\):
     \[
     b_1 = \frac{1}{a_1} = \frac{1}{1} = 1.
     \]
   - For \(m = 2\):
     \[
     b_2 = \frac{1}{a_1} + \frac{1}{a_2} = 1 + \frac{1}{2} = \frac{3}{2}.
     \]
   - For \(m = 3\):
     \[
     b_3 = \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}.
     \]

5. **Prove the general form of \(b_n\):**
   We will prove by induction that:
   \[
   b_n = 2 - \frac{1}{a_{n+1} - 1}.
   \]
   - **Base case:**
     For \(n = 1\),
     \[
     b_1 = 1 = 2 - \frac{1}{a_2 - 1} = 2 - \frac{1}{2 - 1} = 2 - 1 = 1.
     \]
   - **Inductive step:**
     Assume \(b_{n-1} = 2 - \frac{1}{a_n - 1}\). Then,
     \[
     b_n = b_{n-1} + \frac{1}{a_n} = 2 - \frac{1}{a_n - 1} + \frac{1}{a_n}.
     \]
     Simplify the expression:
     \[
     b_n = 2 - \frac{1}{a_n - 1} + \frac{1}{a_n} = 2 - \frac{a_n - 1 + 1}{a_n(a_n - 1)} = 2 - \frac{a_n}{a_n(a_n - 1)} = 2 - \frac{1}{a_n(a_n - 1)}.
     \]
     Since \(a_{n+1} = a_n(a_n - 1) + 1\), we have:
     \[
     b_n = 2 - \frac{1}{a_{n+1} - 1}.
     \]
     Thus, by induction, the formula holds for all \(n\).

6. **Determine the limit of \(b_n\):**
   As \(n \to \infty\), \(a_{n+1} \to \infty\), so \(\frac{1}{a_{n+1} - 1} \to 0\). Therefore,
   \[
   \lim_{n \to \infty} b_n = 2.
   \]

7. **Find the least number \(M\):**
   Since \(b_n = 2 - \frac{1}{a_{n+1} - 1}\) and \(\frac{1}{a_{n+1} - 1} > 0\), we have:
   \[
   b_n < 2 \text{ for all } n.
   \]
   Therefore, the least number \(M\) such that \(\sum_{n=1}^{m} \frac{1}{a_n} < M\) for all \(m \in \mathbb{N}\) is \(M = 2\).

The final answer is \(\boxed{2}\).