The problem is the following:
Given $X$ a random variable with MGF defined by $$\psi_X(t)=\left(\frac{e^t-1}{t}\right)^n$$ deduce that the pdf of $X$ is $$f_X(x)=\sum_{k=0}^n(-1)^k\binom{n}{k}\mathbf{1}_{\{k\le x\}}(x)\frac{(x-k)^{n-1}}{(n-1)!}$$ for $x \in [0,n]$.
Initially, I tried to use the inversion formula on the characteristic function, because $$f(x)=\int_{\mathbb{R}}e ^{-itx} \varphi_X(t)\,\text{d}t=\int_{\mathbb{R}}e^{-itx}\psi_X(it)\,\text{d}t$$ but this seems to go nowhere since the integral that pops out is not very "solvable". In fact: $$\psi_X(t)=\frac{1}{t^n}\sum_{k=0}^n\binom{n}{k}e^{t(n-k)}(-1)^k \Longrightarrow f(x)=\sum_{k=0}^n\binom{n}{k}(-1)^k\int_{\mathbb{R}}\frac{e^{it(n-k-x)}}{t^n}\, \text{d}t$$ And I don't know how to proceed.
Then, I tried with induction on $n$ using the convolution formula in the following way: $$\psi_X(t)=\psi_{X'}(t)\psi_Y(t)=\psi_{X'+Y}(t)\qquad \text{where} \qquad \psi_{X'}(t)=\left(\frac{e^t-1}{t}\right)^{n-1},\ \psi_Y(t)=\frac{e^t-1}{t}$$ So $X'$ and $Y$ are two independent random variables with $Y$ uniform on $(0,1)$ (i.e. $f_Y=\mathbf{1}_{(0,1)}$) and $X'$ with pdf given (by induction hypothesis) by: $$f_{X'}(x)=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\mathbf{1}_{\{k\le x\}}(x)\frac{(x-k)^{n-2}}{(n-2)!}$$ This means that $$f_X(t)=f_{X'+Y}(t)=\int_{\mathbb{R}}f_{X'}(x)f_Y(t-x)\, \text{d}x\\=\sum_{k=0}^{n-1}\frac{(-1)^k}{(n-2)!}\binom{n-1}{k}\int_{\mathbb{R}}(x-k)^{n-2}\mathbf{1}_{\{k\le x\}}(x)\mathbf{1}_{[0,n-1]}(x)\mathbf{1}_{(0,1)}(t-x)\, \text{d}x\\=\sum_{k=0}^{n-1}\frac{(-1)^k}{(n-2)!}\binom{n-1}{k}\int_{\mathbb{R}}(x-k)^{n-2}\mathbf{1}_{[k,n-1]}(x)\mathbf{1}_{(t-1,t)}(x)\, \text{d}x$$ My problem is with those indicator functions, since $\mathbf{1}_{[k,n-1]}\mathbf{1}_{(t-1,t)}=\mathbf{1}_{[k,n-1]\cap (t-1,t)}$ and I don't understand how can I write this so that I have the "right" integration domain. How can I conclude? Is this path correct?
Update: I almost managed to find a full solution using the decomposition: $$\int_{\mathbb{R}}(x-k)^{n-2}\mathbf{1}_{[k,n-1]\cap(t-1,t)}(x)\, \text{d}x=\\ \mathbf{1}_{[k+1,n-1]}(t)\int_{t-1}^t (x-k)^{n-2}\, \text{d}x+\mathbf{1}_{[k,k+1)}(t)\int_{k}^t (x-k)^{n-2}\, \text{d}x+\mathbf{1}_{(n-1,n]}(t)\int_{t-1}^{n-1} (x-k)^{n-2}\, \text{d}x$$ valid for $k \in \{0,\dots,n-2\}$. Now the problem is that, if I suppose $t\le n-1$ the last addend is zero, and straightforward calculations bring to the thesis, but in the case $t \in (n-1,n]$ then the first two addends are zero, and we obtain the sum $$f_X(t)=\sum_{k=1}^{n-1}\frac{(-1)^k}{(n-1)!}\binom{n-1}{k-1}((t-k)^{n-1}-(n-k)^{n-1})$$ And from this sum it seems to me that we cannot obtain the one that we want.
Where is the problem?