Find the value $\sum_{n=1}^{\infty}(e-(1+\frac{1}{n})^n)$

How to find the following series' value?

$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^n\bigg)$$

The sum is diverging as for $n>2$ $$e-\left(1+\frac{1}{n}\right)^n > \frac{1}{n}$$

The telescoping series $$e-\left({1+{1\over n}}\right)^n=\sum_{j=1}^n\left({1+{1\over n}}\right)^{j-1}\left[\exp(1/n)-\left({1+{1\over n}}\right)\right] \exp((n-j)/n)$$ shows that $$e-\left({1+{1\over n}}\right)^n\geq n \left[\exp(1/n)-\left({1+{1\over n}}\right)\right]\geq n \,{1\over 2}\left({1\over n}\right)^2 = {1\over 2n}$$ for all $n\geq 1$. Therefore the OP's series diverges by comparison with the harmonic series.