Here's a direct proof. Let $M$ be a smooth $n$-manifold and $f$ a Morse function on $M$. Let $p_i$ be the local minima, $q_j$ be the index-$1$ critical points, $f(p_i)<f(q_j)$ for all $i,j$, and $\gamma_{ij}$ the gradient flow line from $p_i$ to $q_j$. Connect each $p_i$ to $p_0$ by a direct path of $\gamma_{ij}$s. Note in particular the graph we have defined is contractible. Homotope $f$ so the $q_j$ which are not on any paths have a greater value than the $q_j$ which are on the path, say (by scaling and translating) $f^{-1}(1)$ separates the greater $q_j$ and the $q_j$ on the path. Now consider $B = f^{-1}(\infty,1]$. By construction, this is diffeomorphic to an $n$-ball. In $B$, replace $f$ by the radial distance function (possibly homotoped to match smoothly with $f$ on $\partial B$). This new $f$, call it $\widetilde{f}$, has exactly one index $0$ critical point.
Repeat the procedure for $-\widetilde{f}$ to get exactly one index $n$ critical point.
Intuitively, think of a level surface flow. It starts with a bunch of dots, the index $0$ critical points, expanding. Then they send out tendrils which join together at the index $1$ critical points. Controlling the Morse function is tantamount to controlling the level surface flow. We slow down the level surfaces near some of the index $1$ points so that the level surfaces all join into a giant sphere. This is our motivation; inside the sphere, we modify the function so that the sphere has expanded from a single dot, instead of many dots.
Here's another proof using handlebody decompositions.
By duality between Morse theory and handlebody theory, "Every manifold admits a Morse function with exactly one local maximum and exactly one local minimum" is equivalent to saying that every closed manifold admits a handlebody decomposition with exactly one $0$-handle and exactly one $n$-handle. To see this, take a smooth handlebody decomposition of $M$ ("smooth" so that the attaching maps are all smooth maps). Since this is done by gluing successive handles, we may focus our attention on the $0$-handles $M^0$ and the manifold obtained by gluing $1$-handles, $M^1$.
Since $M$ is connected and the only handle with disconnected attaching spheres are $1$-handles, $M^1$ is connected (that is, the gluing of $1$-handles kills all elements of $\pi_0$). Therefore, we may pick a single $0$-handle and connect it to each other $0$-handle by a path of $1$-handles (possibly through other $0$-handles). This union of $0$- and $1$-handles is homeomorphic to a ball, so we may replace it with a single $0$-handle and regard the remaining $1$-handles as attaching to the single $0$-handle.
Turn the handlebody decomposition upside down and repeat to obtain a single $n$-handle.
References to look into: Milnor's Morse Theory, Milnor's "Killing paper", Ranicki's book on surgery.
