Milnor’s original construction of an exotic $\mathbb{S}^7$ is fairly explicit and it’s reasonable to think that the resulting space is too weirdly twisted to actually be a sphere in the usual sense. Showing that it is not a standard sphere is very non-trivial and one of Milnor's key insights.
However, it's not to hard to cook up a Morse function on this space with only two critical points, and from that you can see that the space has to be a sphere topologically. Intuitively, the space is topologically a sphere but because of the twisting there is no consistent way to identify ``directions" compared to a standard sphere if one travels from the North to South Pole. Here's the basic gist of how this goes.
For a normal sphere, we can consider $\mathbb{B}^4$ as the the unit ball in $\mathbb{R}^4$, and let $\mathbb{S}^3$ be its boundary. The boundary of $\mathbb{B}^4 \times \mathbb{S}^3$ would then be $\mathbb{S}^3 \times \mathbb{S}^3$. We can then take two copies of $\mathbb{B}^4 \times \mathbb{S}^3$ and glue them together along the boundary. Putting a standard flip into the $\mathbb{S}^3$ part and doing the gluing yields a sphere in the usual sense.*
However, if we consider $\mathbb{R}^4$ as the space of quaternions $\mathbb{H}$ and $\mathbb{S}^3$ as the space of unit quaternions. Take two copies of $\mathbb{R}^4 \times \mathbb{S}^3$ and identify the subsets $\left(\mathbb{R}^4-\{0\}\right) \times \mathbb{S}^3$ under the diffeomorphism
$$
(u, v) \rightarrow\left(u^{\prime}, v^{\prime}\right)=\left(\frac{u} {\|u\|^2}, \frac{u^2 v u^{-1}}{\|u\|}\right)
$$
(using quaternion multiplication). Here, we have "twisted" the identification using quaternionic multiplication (and also extended the gluing to not just occur on the boundary). So it is pretty reasonable to think that this space is not a sphere in the usual sense.
But then you might stumble on the function
$$f=\frac{\Re(v)}{\left(1+\|u\|^2\right)^{1 / 2}}=\frac{\Re\left(u^{\prime \prime}\right)}{\left(1+\left\|u^{\prime \prime}\right\|^2\right)^{1 / 2}}$$
where $u^{\prime \prime}=v^{\prime} u^{\prime}$ and $\Re$ indicates the real part of the quaternion.
A bit of effort will show that this is a Morse function with only two critical points, and thus the space must be a sphere in the sense of topology, which doesn't see the smooth structure.
*If you glue the boundaries without the flip, the resulting space is $\mathbb{S}^4 \times \mathbb{S}^3$. Thanks to @HenrikRüping for pointing this out.
