Why is $\mathbb{E}[X] = 1 + \sum^\infty_{k=1}\mathbb{P}(X > k)$ true?

$aP(X = a) = P(X=a) + \dots + P(X = a)$ ($a$ times)

Now assuming $X$ takes only integer values $\ge 1$, you have

$$1 = P(X > 0) = P(X = 1) + \color{red}{ P(X = 2)} + \color{green}{P(X=3) }\dots$$ $$P(X > 1) = \color{red}{P(X = 2)} + \color{green}{P(X = 3) }+ \dots $$ $$P( X > 2) = \color{green}{P(X = 3)} + P(X = 4) + \dots $$

and so on.

Summing everything on the LHS you get $ 1 + \sum_{k=1}^\infty P(X > k)$, on the RHS you find $\sum_{a = 1}^\infty aP(X = a)$

Hence $$E[X] = \sum_{a = 1}^\infty aP(X = a) = 1 + \sum_{k=1}^\infty P(X > k)$$

Let $X$ be a nonnegative random variable with distribution function $F_X$. Then, by Fubini's Theorem, we have $$E[X]=\int_{[0,\infty)}\,x\,\text{d}F_X(x)=\int_{[0,\infty)}\,\int_{[0,\infty)}\,\chi_{[0,x)}(t)\,\text{d}t\,\text{d}F_X(x)=\int_{[0,\infty)}\,\int_{[0,\infty)}\,\chi_{[0,x)}(t)\,\text{d}F_X(x)\,\text{d}t\,,$$ where $\chi_E$ is the characteristic function on a set $E$. That is, $$E[X]=\int_{[0,\infty)}\,\int_{[0,\infty)}\,\chi_{(t,\infty)}(x)\,\text{d}F_X(x)\,\text{d}t=\int_{[0,\infty)}\,\text{Prob}(X> t)\,\text{d}t\,.$$ In particular, if $X$ is discrete, say, $X(\omega)\in\left\{x_1,x_2,\ldots\right\}$ for all $\omega$ with $0\leq x_1<x_2<\ldots$ and $\lim\limits_{n\to\infty}\,x_n=\infty$ (this ordering doesn't always exist, for example, when the possible values of $X$ are in $\mathbb{Q}_{>0}$), then $$\begin{align} E[X]&=\int_{[0,x_1)}\,\text{Prob}(X>t)\,\text{d}t+\sum_{i=1}^\infty\,\int_{[x_{i},x_{i+1})}\,\text{Prob}(X>t)\,\text{d}t \\&=\int_{[0,x_1)}\,1\,\text{d}t+\sum_{i=1}^\infty\,\int_{[x_{i},x_{i+1})}\,\text{Prob}(X>x_{i})\,\text{d}t \\ &=x_1+\sum_{i=1}^\infty\,\left(x_{i+1}-x_i\right)\,\text{Prob}\left(X>x_{i}\right)\,. \end{align}$$ If $\left\{x_1,x_2,\ldots\right\}=\{1,2,\ldots\}$, then we get $$E[X]=1+\sum_{i=1}^\infty\,\text{Prob}(X>i)\,,$$ as required. As Zachary Selk mentioned, if $\left\{x_1,x_2,\ldots\right\}=\{0,1,2,\ldots\}$, then $$E[X]=\sum_{i=0}^\infty\,\text{Prob}(X>i)\,.$$