Let us consider a principal bundle \(P \overset{\pi }{\to} M\), let \(u \in P\) and let \(G_p\) be the fibre at \(p = \pi(u)\). According to my book, the vertical subspace \(V_uP\) is defined as the subspace of \(T_uP\) which is tangent to \(G_p\) at \(u\). Subsequently, let \(A \in \mathfrak{g}\), where \(\mathfrak{g}\) denotes the Lie algebra, then the fundamental vector \(A^\sharp \in T_uP\) is defined by:

\(A^\sharp f(u) = \frac{\mathrm{d}}{\mathrm{d} t} f (u \exp (t A))|_{t=0}\)

where \(f: P \to \mathbb{R}\) is an arbitrary smooth function. Now the book states:

The vector \(A^\sharp\) is tangent to \(P\) at \(u\), hence \(A^\# \in V_uP\).

Why does \(A^\sharp \in V_uP\)?

P.S. I do understand that since we are considering a principal bundle, we have:

\(\pi(u) = \pi(u \exp (t A)) = p\)

and so \(u \exp (t A) \in G_p\), but I'm not sure how/if this is helpful.