What's the kurtosis of exponential distribution?

Often "kurtosis" is taken to mean "excess kurtosis", i.e. the amount by which the kurtosis exceeds that of the normal distribution, thus the kurtosis minus $3.$

Subtraction of $3$ makes sense in some contexts even without thinking about the normal distribution. Let $\mu=\operatorname{E}(X)$ and note that the two functionals $$ \operatorname{A}(X) = \operatorname{E}\big((X-\mu)^4\big) \quad \text{and} \quad \operatorname{B}(X) = \Big(\operatorname{E}\big((X-\mu)^2\big) \Big)^2 $$ are $(1)$ homogenous of degree $4$ (i.e. multiplying $X$ by a scalar $c$ multiplies the value of the functional by $c^4,$ and $(2)$ translation-invariant. But they are not "cumulative", i.e. for independent random variables $X_1,\ldots,X_n$ we do not have $\operatorname{A}(X_1+\cdots+X_n) = \operatorname{A}(X_1)+\cdots+\operatorname{A}(X_n)$ nor $\operatorname{B}(X_1+\cdots+X_n) = \operatorname{B}(X_1)+\cdots+\operatorname{B}(X_n).$ But $\kappa = {\operatorname{A}}-{3\operatorname{B}}$ is homogenous of degree $4$ and translation invariant and cumulative. And for any coefficient besides $-3$ that doesn't work. This quantity $\kappa(X)$ is the fourth cumulant of the distribution of $X.$ The excess kurtosis is $$ \frac{\kappa(X)}{\sigma^4}. $$

A more careful reading of the MathWorld article states that the kurtosis excess of an exponential distribution is $6$. This is not the same as kurtosis, since the former is defined as $$\gamma_2 = \frac{\mu_4}{\mu_2^2} - 3,$$ (see Equation 3), as opposed to the simple kurtosis $$\beta_2 = \frac{\mu_4}{\mu_2^2},$$ which is Equation (1). The reason for these two definitions is also explained in the article, since $\beta_2$ for a normal distribution is $3$, hence the definition of kurtosis excess representing in some sense the amount of kurtosis in excess of a normal distribution.

Therefore, there is no error. MathWorld is correct.

I get for $\lambda=1$ $$\int_0^\infty(x-1)^4\exp(-x)\,dx=9.$$ In general I get $$\int_0^\infty(x-1)^n\exp(-x)\,dx=D_n,$$ the $n$-th derangement number.

Wikipedia is correctable.