Step 1: Note the helpful rule.
Two angles that add up to $90^\circ$ are called complementary. For them, \[ \tan\theta = \cot(90^\circ - \theta) \]
Step 2: Turn this into a product.
Since $\cot\alpha = \dfrac{1}{\tan\alpha}$, multiplying a tangent by the tangent of its complement gives \[ \tan\theta \cdot \tan(90^\circ - \theta) = 1 \]
Step 3: Pair the terms from the ends.
In $\tan 1^\circ \cdot \tan 2^\circ \cdots \tan 89^\circ$, pair the first with the last, the second with the second-last, and so on. For example $\tan 1^\circ \cdot \tan 89^\circ$.
Step 4: Evaluate each pair.
Each pair adds to $90^\circ$, so each pair equals $1$. \[ \tan 1^\circ \cdot \tan 89^\circ = 1, \quad \tan 2^\circ \cdot \tan 88^\circ = 1, \ \ldots \]
Step 5: Handle the middle term.
The only term left alone is the middle one, $\tan 45^\circ = 1$.
Step 6: Multiply everything.
All pairs give $1$ and the middle term is $1$, so the whole product is $1$. \[ \boxed{1} \]