Given that $(1 + \tan 1^{\circ})(1 + \tan 2^{\circ}) \ldots (1 + \tan 45^{\circ}) = 2^n$, find $n$.
1. We start by considering the product \((1 + \tan 1^{\circ})(1 + \tan 2^{\circ}) \ldots (1 + \tan 45^{\circ})\).

2. Notice that for any pair of angles \(\alpha\) and \(\beta\) such that \(\alpha + \beta = 45^{\circ}\), we have:
   \[
   (1 + \tan \alpha)(1 + \tan \beta)
   \]
   Expanding this, we get:
   \[
   1 + \tan \alpha + \tan \beta + \tan \alpha \tan \beta
   \]

3. Using the tangent addition formula for \(\alpha + \beta = 45^{\circ}\):
   \[
   \tan(\alpha + \beta) = \tan 45^{\circ} = 1
   \]
   We know:
   \[
   \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 1
   \]
   Therefore:
   \[
   1 = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}
   \]
   Solving for \(\tan \alpha \tan \beta\):
   \[
   1 - \tan \alpha \tan \beta = \tan \alpha + \tan \beta
   \]
   Rearranging, we get:
   \[
   \tan \alpha + \tan \beta + \tan \alpha \tan \beta = 1
   \]
   Adding 1 to both sides:
   \[
   1 + \tan \alpha + \tan \beta + \tan \alpha \tan \beta = 2
   \]
   Thus:
   \[
   (1 + \tan \alpha)(1 + \tan \beta) = 2
   \]

4. We have 22 such pairs of angles: \((1^{\circ}, 44^{\circ}), (2^{\circ}, 43^{\circ}), \ldots, (22^{\circ}, 23^{\circ})\).

5. Additionally, we have the term for \(45^{\circ}\):
   \[
   1 + \tan 45^{\circ} = 1 + 1 = 2
   \]

6. Therefore, the entire product is:
   \[
   (1 + \tan 1^{\circ})(1 + \tan 2^{\circ}) \ldots (1 + \tan 45^{\circ}) = 2^{22} \times 2 = 2^{23}
   \]

Conclusion:
\[
\boxed{23}
\]