Let the distribution of card shuffling describe the order of cards in a deck. And let the uniform distribution the distribution of which every permutation in has the same chance. prove that the maximum difference (by probability )of uniform distribution and taking card atop and putting it randomly in the deck is $1-1/(n-1)!$. This claim is made in "proofs from the book" page 222.
Link to the book: http://cslabcms.nju.edu.cn/problem_solving/images/b/b3/Proofs_from_THE_BOOK_%28Fifth_Edition_2014%29.pdf
In the linked book, page 208 the starting distribution before "taking the card atop and putting it randomly in the deck" was identity distribution, that is cards were numbered $1,2,\ldots,n$, from top to the bottom respectively. Now let $Q$ be the distribution of the cards after taking the card atop and putting it randomly in the deck. Also, let $U$ be the uniform distribution of the cards. Then the total variation distance between them is
$$ d_{TV}(Q,U) = \frac{1}{2}\sum_{\pi \in \sigma_n}|Q(\pi)-U(\pi)|, $$ where $\sigma_n$ is the set of all permutations of $\{1,2,\ldots,n\}$. Now note that $Q$ puts mass $\frac{1}{n}$ on only $n$ elements in $\sigma_n$ namely $12\ldots n,2134\ldots n,\ldots,234\ldots n1$, let us call support of $Q$ to be $s_n$. Now $$ d_{TV}(Q,U) = \frac{1}{2}\sum_{\pi \in s_n}|Q(\pi)-U(\pi)|+ \frac{1}{2}\sum_{\pi \in \sigma_n\setminus s_n}|Q(\pi)-U(\pi)|\\ = \frac{1}{2}\sum_{\pi \in s_n}\left|\frac{1}{n}-\frac{1}{n!}\right|+ \frac{1}{2}\sum_{\pi \in \sigma_n\setminus s_n}\left|0-\frac{1}{n!}\right|\\ =\frac{1}{2}n\left|\frac{1}{n}-\frac{1}{n!}\right|+ \frac{1}{2}(n!-n)\left|0-\frac{1}{n!}\right|. $$ Now simplify to get the answer.