Let $$T(z)=\sum_{1}^\infty \frac{n^{n-1}x^n}{n!}.$$ This is known as the exponential generating function of unordered rooted trees, see for example https://math.berkeley.edu/~mhaiman/math172-spring10/trees.pdf. This function solves $$T(x)=xe^{T(x)}.$$ Your sum is $y=T(1/e)$. So we have to solve the equation $$y=e^{y-1}.$$ This has a root $y=1$ which is multiple (of multiplicity $2$). This means that the graph of the LHS is tangent to the graph of the RHS at the point $(1,1)$. Since the LHS is linear and RHS is convex, our equation has unique solution, namely $y=1$, which proves your formula.

Let $$T(z)=\sum_{1}^\infty \frac{n^{n-1}x^n}{n!}.$$ This is known as the exponential generating function of unordered rooted trees, It follows from the Burmann-Lagrange formula (example 1 on p. 3) that this function solves $$T(x)=xe^{T(x)}.$$ Your sum is $y=T(1/e)$. So we have to solve the equation $$y=e^{y-1}.$$ This has a root $y=1$ which is multiple (of multiplicity $2$). This means that the graph of the LHS is tangent to the graph of the RHS at the point $(1,1)$. Since the LHS is linear and RHS is convex, our equation has unique solution, namely $y=1$, which proves your formula.

Let $$T(z)=\sum_{1}^\infty \frac{n^{n-1}x^n}{n!}.$$ This is known as the exponential generating function of rooted labeled trees, see for example https://math.berkeley.edu/~mhaiman/math172-spring10/trees.pdf. This function solves $$T(x)=xe^{T(x)}.$$ Your sum is $y=T(1/e)$. So we have to solve the equation $$y=e^{y-1}.$$ This has a root $y=1$ which is multiple (of multiplicity $2$). This means that the graph of the LHS is tangent to the graph of the RHS at the point $(1,1)$. Since the LHS is linear and RHS is convex, our equation has unique solution, namely $y=1$, which proves your formula.

Let $$T(z)=\sum_{1}^\infty \frac{n^{n-1}x^n}{n!}.$$ This is known as the exponential generating function of unordered rooted trees, see for example https://math.berkeley.edu/~mhaiman/math172-spring10/trees.pdf. This function solves $$T(x)=xe^{T(x)}.$$ Your sum is $y=T(1/e)$. So we have to solve the equation $$y=e^{y-1}.$$ This has a root $y=1$ which is multiple (of multiplicity $2$). This means that the graph of the LHS is tangent to the graph of the RHS at the point $(1,1)$. Since the LHS is linear and RHS is convex, our equation has unique solution, namely $y=1$, which proves your formula.