For an arbitrary square matrix $M$, define
$$\exp(M)=I+\frac M{1!}+\frac{M^2}{2!}+\frac{M^3}{3!}+\ldots.$$Construct $2\times2$ matrices $A$ and $B$ such that $\exp(A+B)\ne\exp(A)\exp(B)$.
To construct \(2 \times 2\) matrices \(A\) and \(B\) such that \(\exp(A+B) \ne \exp(A)\exp(B)\), we need to find matrices \(A\) and \(B\) that do not commute, i.e., \(AB \ne BA\). 

1. **Choose non-commuting matrices**:
   Let's choose the following matrices:
   \[
   A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}
   \]
   These matrices do not commute because:
   \[
   AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
   \]
   Clearly, \(AB \ne BA\).

2. **Compute \(\exp(A)\)**:
   The matrix exponential \(\exp(A)\) is given by:
   \[
   \exp(A) = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots
   \]
   Since \(A^2 = 0\), all higher powers of \(A\) are zero. Thus:
   \[
   \exp(A) = I + A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
   \]

3. **Compute \(\exp(B)\)**:
   Similarly, for \(B\):
   \[
   \exp(B) = I + B + \frac{B^2}{2!} + \frac{B^3}{3!} + \cdots
   \]
   Since \(B^2 = 0\), all higher powers of \(B\) are zero. Thus:
   \[
   \exp(B) = I + B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
   \]

4. **Compute \(\exp(A)\exp(B)\)**:
   \[
   \exp(A)\exp(B) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 1 & 0 \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}
   \]

5. **Compute \(A + B\)**:
   \[
   A + B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
   \]

6. **Compute \(\exp(A+B)\)**:
   The matrix \(A + B\) is a rotation matrix, and its exponential can be computed using the formula for the exponential of a 2x2 matrix:
   \[
   \exp(A+B) = \cosh(1)I + \sinh(1)(A+B)
   \]
   where \(I\) is the identity matrix. For simplicity, we can use the fact that:
   \[
   \exp(A+B) = \begin{pmatrix} \cosh(1) & \sinh(1) \\ \sinh(1) & \cosh(1) \end{pmatrix}
   \]

7. **Compare \(\exp(A+B)\) and \(\exp(A)\exp(B)\)**:
   Clearly, \(\exp(A+B) = \begin{pmatrix} \cosh(1) & \sinh(1) \\ \sinh(1) & \cosh(1) \end{pmatrix}\) is not equal to \(\exp(A)\exp(B) = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\).

Thus, we have constructed matrices \(A\) and \(B\) such that \(\exp(A+B) \ne \exp(A)\exp(B)\).

The final answer is \( \boxed{ \exp(A+B) \ne \exp(A)\exp(B) } \).