Let's assume that an equation-free intuition is not possible, and still insist on boiling down the math to the very essentials to get an idea of what's going on: we are trying to obtain the [statistical moments](https://en.wikipedia.org/wiki/Moment_(mathematics)), which, after the obligatory reference to [physics](https://en.wikipedia.org/wiki/Moment_(physics)), we define as **the expected value of a power of a random variable.** For a continuous random variable, the raw $k$-th moment is by LOTUS:

\begin{align}\large \color{red}{\mathbb{E}[{X^k}]} &= \displaystyle\int_{-\infty}^{\infty}\color{blue}{X^k}\,\,\color{green}{\text{pdf}}\,\,\,dx.\end{align}

The moment generating function, $M_X(t)$, is a **way to walk around this integral** by, instead, carrying out:

\begin{align} \large\mathbb{E}[\color{blue}{e^{\,tX}}]&=\displaystyle \int_{-\infty}^{\infty}\color{blue}{e^{tX}}\,\color{green}{\text{pdf}}\, dx.\end{align}



Why? Because it's easier and there is a fantastic property of the MGF that can be seen by expanding the [Maclaurin series](https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions) of $\color{blue}{e^{\,tX}}$ within the expectation operator:

$$M_X(t) = 1 +  \frac{\color{red}{\mathbb{E} \left[X\right]}}{1!} \, t  \, +  \frac{\color{red}{\mathbb{E} \left[X^2\right]}}{2!} \, t^2  \, +  \frac{\color{red}{\mathbb{E} \left[X^3\right]}}{3!} \, t^3 \, + \cdots$$

namely, the moments appear "perched" on this polynomial "clothesline", ready to be culled by simply differentiating $k$ times and evaluating at zero once we go through the easier integration **just once for all moments!**

The fact that it is an easier integration is most apparent when the pdf is an exponential.