The Wikipedia piece on generalized hypergeometric functions lists a reduction formula that looks suspicious to me. Just over Euler's integral transform (link):
$$ {}_{A+1}F_{B+1}\left[ \begin{array}{c} a_{1},\ldots ,a_{A},c + n \\ b_{1},\ldots ,b_{B},c \end{array} ;z\right] = \sum_{j = 0}^n \binom{n}{j} \frac{1}{(c)_j} \frac{\prod_{i = 1}^A (a_i)_j}{\prod_{i = 1}^B (b_i)_j} {}_AF_B\left[ \begin{array}{c} a_{1} + j,\ldots ,a_{A} + j \\ b_{1} + j,\ldots ,b_{B} + j \end{array} ;z\right] $$
Applying this rule to $(z+1)\exp(z) = {}_1 F_1(2;1;z)$ I get $2\exp(z)$.
There is a reference for the formula to a 1990 Collection that doesn't seem to be available online.
Could anyone (a) confirm that the equation is faulty in its current version on Wikipedia and (b) supply a correct Version?
Indeed the expression on the Wikipedia page has an error. As your example suggests, the right hand side is missing some factors of $z$. A more general version of the order reduction formula can be found here. Using the special case of that formula where we reduce the order only by $1$, the correct version of the Wikipedia formula should be $$ {}_{A+1}F_{B+1}\left[ \begin{array}{c} a_{1},\ldots ,a_{A},c + n \\ b_{1},\ldots ,b_{B},c \end{array} ;z\right] = \sum_{j = 0}^n \binom{n}{j} \frac{1}{(c)_j} \frac{\prod_{i = 1}^A (a_i)_j}{\prod_{i = 1}^B (b_i)_j} \color{red}{z^j} {}_AF_B\left[ \begin{array}{c} a_{1} + j,\ldots ,a_{A} + j \\ b_{1} + j,\ldots ,b_{B} + j \end{array} ;z\right] $$