These Numerators Diverge Too

So $\dfrac0{0!} + \dfrac1{1!} + \dfrac2{2!} + \dfrac3{3!} + \dfrac4{4!} + \cdots$

can be rewritten as $\dfrac0{0!} + \dfrac1{1!} + \dfrac2{2*1} + \dfrac3{3*2!} + \dfrac4{4*3!} + \cdots$

$\implies \dfrac0{0!} + \dfrac1{1!} + \dfrac{\color{#D61F06}{\cancel{\color{#333333}{2}}}}{\color{#D61F06}{\cancel{\color{#333333}{2}}}*1} + \dfrac{\color{#D61F06}{\cancel{\color{#333333}{3}}}}{{\color{#D61F06}{\cancel{\color{#333333}{3}}}}*2!} + \dfrac{\color{#D61F06}{\cancel{\color{#333333}{4}}}}{{\color{#D61F06}{\cancel{\color{#333333}{4}}}}*3!} + \cdots$

$\implies \dfrac0{0!} + \dfrac1{0!} + \quad \dfrac1{1!} \quad + \quad \dfrac1{2!} \quad + \quad\dfrac1{3!}\quad + \cdots \qquad \qquad \color{#20A900}{\small\text{Note: 0! = 1, which also meant 0!=1!}}$

Which is the same thing as $e = \dfrac1{0!} + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots$

Therefore, the answer is $\color{#302B94}{\boxed{\Huge {\text{True}}}}$

$\begin{aligned} e^x & = \sum_{k=0}^\infty \frac {x^k}{k!} \\ \frac d{dx} e^x & = \frac d{dx} \sum_{k=0}^\infty \frac {x^k}{k!} \\ e^x & = 0 + \sum_{k=1}^\infty \frac {kx^{k-1}}{k!} & \small \color{#3D99F6}{\text{Putting }x=1} \\ \implies \boxed{e} & = \frac 0{0!} + \frac 1{1!} + \frac 2{2!} + \frac 3{3!} + \frac 4{4!} + ... \end{aligned}$